Quantitative Aptitude Questions Bank PO Tricks for SBI PO, IBPS PO, NIACL, NICl, BoB, Dena Bank PGDBF and other competitive exams

1. If 75 is added to 75% of a number then the number becomes 125% of it. Find the original number.
   A) 187.5
   B) 150
   C) 200
   D) 112.5
   E) 165.5
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   75% = 3/4, 125% = 5/4
   We have to make 3/4 to 5/4
   3 => 5 means increase of 2
   So 2 = 75 (the number that is to be added
   And 4 (denominator – the original number)  = 75/2 * 4 = 150[/su_spoiler]
2. After a discount of 11.11%, a trader still makes a gain of 14.28%. At how much percent above the cost price does he mark his goods?
   A) 16.67%
   B) 23.37%
   C) 22.22%
   D) 28.56%
   E) 31.22%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   Discount is calculated on MP
   11.11% = 1/9  (1 – discount, 9 – MP)
   MP – Discount = SP
   So 9 – 1 = 8
   Profit% = 14.28% = 1/7 (1 – gain, 7 – CP)
   SP = CP + Profit
   So SP = 7 + 1 = 8
   Both SP are found to be same
   So CP : SP : MP = 7 : 8 : 9
   Required % = (9-7)/7 * 100 = 200/7% [/su_spoiler]
3. A water tank has three taps A, B, and C. A fills four buckets in 24 mins, B fills 8 buckets in 1 hour and C fills 2 buckets in 20 minutes. If all the taps are opened together a full tank is emptied in 2 hours. If a bucket can hold 5 litres of water, what is the capacity of the tank?
   A) 240 l
   B) 280 l
   C) 200 l
   D) 300 l
   E) 250 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Bucket is to be emptied in 2 hours or 120 minutes
   In 24 mins, A can fill 4 buckets
   So in 120 minutes, it can fill 20 buckets
   Similarly
   In 120 minutes, B can fill 16 buckets
   In 120 minutes, C can fill 12 buckets
   Total = 20+16+12 = 48
   1 bucket = 5 litre
   So 48 buckets = 240 litres [/su_spoiler]
4. 7 men, 5 women and 8 children together have to distribute 1740 books in a total 3 days. On the first day all distribute books. On second day 2 women and 3 children were absent and on the third day 3 men and 5 children were absent. If the ratio of the number of books distributed in a day by a man, a woman and a child was 5 : 4 : 2 respectively, a total of approximately how many books were distributed on the second day?
   A) 650
   B) 700
   C) 510
   D) 600
   E) 570
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   The ratio 5 : 4: 2 specifies the efficiency
   In 3 days, total men = 7 + 7 + 4 = 18
   In 3 days, total women = 5 + 3 + 5 = 13
   In 3 days, total children = 8 + 5 + 3 = 16
   So 18*5x + 13*4x + 16*2x = 1740
   Solve, x = 10
   So second day distribution:
   7*5x + 3*4x + 5*2x = 57x = 57*10 = 570  [/su_spoiler]
5. Keshav invests some amount. It amounts to Rs 6000 in 2 years and to Rs 8000 in 4 years if compounded annually. Fin the sum invested by him.
   A) Rs 4000
   B) Rs 5400
   C) Rs 5000
   D) Rs 4500
   E) Rs 4800
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   When difference of year is same in CI (like here 2 years and then 2+2 = 4 years):
   x(principal) : 6000 = 6000 : 8000
   Solve, x = Rs 4500 [/su_spoiler]
6. The price of an article is first increased by 25%, then by 10% and at last increased by 14 2/7%. If the original price was Rs 140, find the new price after 3 increments.
   A) 180
   B) 160
   C) 200
   D) 230
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   25% = 1/4
   10% = 1/10
   14 2/7% = 1/7
   Original price = 140
   After first increment = 140 + 1/4 * 140 = 175
   After second increment = 175 + 1/10 * 175 = 192.5
   After third increment = 192.5 + 1/7 * 192.5 = Rs 220   [/su_spoiler]
7. A shopkeeper bought an article for Rs 12000. He marked the price 25% above the cost price. If during selling a discount of 10% is given on the marked price, find the profit/loss% incurred by the shopkeeper.
   A) 15%
   B) 12.5%
   C) 12%
   D) 10%
   E) 14.5%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   MP is 25% above CP
   25% = 1/4 (1 – increased price, 4 – CP)
   So MP = CP + increased price = 4(CP) + 1 = 5(MP)  ……..(i)
   Discount = 10% = 1/10
   So SP = 10(MP)  – 1 = 9(SP)  …….(ii)
   Now make MP same in both equations by multiplying equation(i) by 2
   So 8(CP) + 2 = 10 (MP)
   So
   CP : MP : SP = 8 : 10 : 9
   So profit% = (9-8)/8 * 100 = 12.5%
   [/su_spoiler]
8. A solution of sugar syrup has 35% of sugar. Another solution has 20% sugar. How many litres of the second solution must be added to 40 litres of the first solution to make a solution of 25% sugar?
   A) 40 l
   B) 60 l
   C) 80 l
   D) 70 l
   E) 50 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option C
   Solution:
   35% of 40 + 20% of x = 25% of (40+x)   [/su_spoiler]
9. Train A crosses a pole and a platform in 18 seconds and 39 seconds respectively. The length of the platform is 157.5 m. What will be the length of train B if it is equal to the sum of half of the length of train A and twice the length of the platform?
   A) 328.5 m
   B) 382.5 m
   C) 238.5 m
   D) 315 m
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   (l+ 157.5)/s =  39
   Also l/s = 18
   Solve both, l = length of train = 135m
   So length of train B = (135/2) + 2 * 157.5 = 382.5 m [/su_spoiler]
10. A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time, it had to increase its speed by 250km/hr from its usual speed. Find its usual speed.
    A) 720 km/hr
    B) 750 km/hr
    C) 790 km/hr
    D) 840 km/hr
    E) None of these
    [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
    Solution:
    Solve with options:
    Take 750. It cancels out 1500, so 1500/750 = 2 hours
    it left 30 minutes later, so it has to cover distance in 2 – 1/2 = 1 hour 30 minutes  = 3/2 hours
    then speed should be 1500/(3/2) = 1000 km/hr
    Speed is increased by 250 km/hr
    And 750 + (250) = 1000 – satisfies the condition. [/su_spoiler]

Directions: What will come in place of (?) in following series following a certain pattern?

1. 375, 75, 30, 18, ?, 14.4
   A) 15
   B) 15.2
   C) 16
   D) 14.4
   E) 16.5
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Pattern:
   Here last value is in decimals
   So there is a possibility that the numbers are multiplied or divided by some decimal quantity
   × 0.2, × 0.4, × 0.6, × 0.8, × 1.0 [/su_spoiler]
2. 25, 89, 332, 588, 713, ?
   A) 855
   B) 856
   C) 749
   D) 860
   E) 785
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Pattern:  
   When the number increase suddenly, we use multiplication to find pattern. But here they are not increasing suddenly, so try addition/subtraction
   25 + 2⁶, 89 + 3⁵, 332 + 4⁴, 588 + 5³, 713 + 6²
   [/su_spoiler]
3. 18, 36, 62, 96, 138, ?
   A) 172
   B) 206
   C) 244
   D) 220
   E) 188
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Pattern:
   18 + 18 = 36
   36 + (18+8) = 36 + 26 = 62
   62 + (26+8) = 62 + 34 = 96
   96 + (34+8) = 96 + 42 = 138
   138 + (42+8) = 188[/su_spoiler]
4. 16, 24, 60, 210, 945, ?
   A) 5197.5
   B) 4234.5     
   C) 5848.5
   D) 4113.5
   E) 5434.5
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Pattern:
   16×1.5, 24×2.5, 60×3.5, 210×4.5, 945×5.5 [/su_spoiler]
5. 8, 18, 86, 308, 828, ?
   A) 1955
   B) 1902
   C) 1406
   D) 1838
   E) 1629
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Pattern:
   8 + (2^3 + 2) = 18,
   18 + (4^3 + 4) = 86,
   86 + (6^3 + 6) = 308,
   308 + (8^3 + 8) = 828,
   828 + (10^3 + 10) = 1838 [/su_spoiler]
6. 16, 20, 28, 27, 42, ?
   A) 26
   B) 45
   C) 67
   D) 42
   E) 32
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Pattern:
   16………. 20……….. 28…………. 27………. 42……….. 32
   .     +4          +8              -1             +15          -10       
   .          +2²            -3²            +4²            -5² [/su_spoiler]
7. 5, 7, 11, 37, ?, 721
   A) 143
   B) 294
   C) 138
   D) 147
   E) 256
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Pattern:
   5*1 + 2, 7*2 – 3, 11*3 + 4, 37*4 – 5, 143*5 + 6 [/su_spoiler]
8. 21, 33, 85, 301, 1359, ?
   A) 6066
   B) 7359
   C) 7480
   D) 6242
   E) 5155
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Pattern:
   ×1.5 + 1.5, ×2.5 + 2.5,  ×3.5 + 3.5,  ×4.5 + 4.5, ×5.5 + 5.5,        
   [/su_spoiler]   
9. 6, 9, 24, 90, 432, ?
   A) 2380
   B) 1340
   C) 2460
   D) 2520
   E) 1420
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Pattern:
   6×2 – 6/2 = 9,
   9×3 – 9/3 = 24,
   24×4 – 24/4 = 90,
   90×5 – 90/5 = 432,
   432×6 – 432/6 = 2520 [/su_spoiler]
10. 4, 13, 41, 126, ?, 1151
    A) 382
    B) 688
    C) 453
    D) 562
    E) 436
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
    Pattern:
    4×3 + 1 = 13,
    13×3 + 2 = 41,
    41×3 + 3 = 126, 
    126×3 + 4 = 382, 
    382×3 + 5 = 1151  [/su_spoiler]

Bank PO Quantitative Aptitude Questions Answers

1. Fresh apples contain 70% water by weight, whereas dried apples contain 4% water by weight. How many kg of dried apples can be obtained from 24 kg of fresh apples?
   A) 9.3 kg
   B) 7.1 kg
   C) 8.2 kg
   D) 7.5 kg
   E) 9.8 kg
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   In 24 kg of apples,  water is 70%, so pulp is 30% = 30% of 24kg = 7.2 kg
   Let total of x kg dried apples is to be obtained from 24 kg of fresh apples.
   So in this x kg dried apples, 96% should be pulp to have 4% water. So
   96% of x = 7.2 kg
   Solve, x = 7.5 kg[/su_spoiler]
2. There are 420 students residing in a hostel. If the students increase by 100, the expenses of the mess increase by Rs 500 per month, while the average expenditure per head diminishes by Rs 25. What is the original monthly expense of the mess?
   A) Rs 105
   B) Rs 120
   C) Rs 135
   D) Rs 180
   E) Rs 165
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   Let original monthly expense of the mess = Rs x
   So 520 (x – 25) – 420x = 500  [520 is (420+100)]
   Solve, x = Rs 135[/su_spoiler]
3. Points A and B are 45 km apart. Boat P can cover downstream from A to B and back from B to A in a total of 20 hours. Another boat Q can cover the same distance in a total of 8 hours. If the speed of boat P is half the speed of boat Q, then what is the speed of current?
   A) 3 km/hr
   B) 8 km/hr
   C) 6 km/hr
   D) 10 km/hr
   E) 12 km/hr
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Speed of boat Q in still water = 2x km/hr, then of boat P = x km/hr.
   Let speed of current = y km/hr
   So
   45/(x + y)  + 45/(x – y)  = 20
   And also
   45/(2x + y)  + 45/(2x – y)  = 8
   Solve both equations, y = 3 km/hr[/su_spoiler]
4. On selling rice for Rs 120 per kg, a shopkeeper incurs a loss of 10%. What will be the amount of rice sold, if there was a net loss of Rs 40?
   A) 5 kg
   B) 2.5 kg
   C) 4 kg
   D) 3 kg
   E) 4.5 kg
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   10% of total CP = Rs 40
   So 100% = Rs 400, means total CP = Rs 400
   So SP = 400 – 40 = Rs 360
   So total rice sold at this price = 360/120 = 3kg  [/su_spoiler]
5. Can A contains 108 l of milk of milk and water in ratio 7 : 1. Another can B contains milk and water in ratio 6 : 1. If the mixtures of both the cans are mixed in another can, the respective ratio becomes 13 : 2. Find the amount of new mixture formed?
   A) 94.5 l
   B) 119.5 l
   C) 143.5 l
   D) 187.5 l
   E) 202.5 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   By rule of allegation:
   (108 l)………..…………..(x l)
   1/8………………………1/7
   …………..……2/15
   1/7*15………………….1/8*15
   So ratio is 1/7*15 : 1/8*15 = 8 : 7
   So 108/x = 8/7
   Solve, x = 94.5 l
   So total mixture formed is 108 + 94.5 = 202.5 l [/su_spoiler]
6. A letter is chosen at random from ‘ASPIRANTSZONE’. What is the probability that it is a vowel?
   A) 4/13
   B) 8/13
   C) 6/13
   D) 7/13
   E) 5/13
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   There are total 13 letters in ASPIRANTSZONE
   And total 5 vowels – 2 A’s, 1 I, 1 E, 1 O
   So probability is 5/13 [/su_spoiler]
7. A shopkeeper bought rice at the rate of 12 Rs/kg. While selling he uses false weights of 750 gms instead of 1 kg. If he sells rice making a profit of 60%, then he sold rice at what rate (per kg)?
   A) Rs 16.9
   B) Rs 15.7
   C) Rs 12.8
   D) Rs 14.4
   E) Rs 15.3
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   Solution:
   When he uses 750 gms instead of 1000 gms. He makes a profit of
   (1000-750)/750 × 100 = 100/3%
   Let on CP he gains x%.
   So using successive formula:
   100/3 + x + (100/3 * x)/100 = 60
   Solve x = 20 %
   Now CP is Rs 12. Profit % is 20 %
   Find SP.
   SP = 120% of 12 = Rs 14.4 per kg.
   [/su_spoiler]
8. A sum of Rs 6800 is deposited in two schemes. One part is deposited in Scheme A which offers 12% rate of interest. Remaining part is invested in Scheme B which offers 10% rate of interest compounded annually. If interest obtained in scheme A after 5 years is Rs 678 more than the interest obtained in scheme B after 2 years, find the part deposited in scheme A.
   A) Rs 4200
   B) Rs 3900
   C) Rs 3400
   D) Rs 2600
   E) Rs 3000
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option D
   Solution:
   (6800-x)*12*5/100 = x [ (1 + 10/100)² – 1] + 678
   68*12*5 – 60x/100 = 21x/100 + 678
   Solve, x = Rs 4200
   So in scheme A = 6800-4200 = Rs 2600[/su_spoiler]
9. Ratio of age of A 3 years hence to C 4 years hence is 6 : 7. Age of B is 9 years greater than the age of A and the ratio of age of B 2 years ago to that of C 8 years ago is 4 : 3. Find the sum of all of A, B and C’s present ages.
   A) 134
   B) 129
   C) 124
   D) 108
   E) 113
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]       Option E
   Solution:
   (A+3)/(C+4) = 6/7
   B = 9 + A
   (B-2)/(C-8) = 4/3
   Solve all, A = 33, B = 42,  = 38[/su_spoiler]
10. A and B started a business by investing Rs 2400 and Rs 2000 respectively. After 6 months A added Rs 900 while B added Rs 700 to their respective investments. If after a year, out of total profit share of A is Rs 2500 more than the share of B, find the total profit.
    A) Rs 34000
    B) Rs 25000
    C) Rs 29000
    D) Rs 31000
    E) Rs 26000
    [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
    Solution:
    Ratio of shares A : B is
    2400*6 + 3300*6 : 2000*6 + 2700*6
    24 + 33 : 20 + 27
    57 : 47
    So (57-47)/(57+47) * x = 2500
    Solve, x = Rs 26000 [/su_spoiler]

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

1. I. 3x² + 22x + 24 = 0,
   II. 3y² – 8y – 16 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   3x² + 22 x + 24 = 0
   3x² + 18x + 4x + 24 = 0
   Gives x = -4/3, -6
   3y² – 8y – 16 = 0
   3y² – 12y + 4y – 16 = 0
   So y = -4/3, 4
   Plot on number line
   -6…. -4/3……. 4[/su_spoiler]
2. I. 5x² – 18x – 8 = 0,
   II. 2y² + 11y + 12 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   5x² – 18x – 8 = 0
   5x² – 20x + 2x – 8 = 0
   So x = -2/5, 4
   2y² + 11y + 12 = 0
   2y² + 8y + 3y + 12 = 0
   Gives y = -4, -3/2
   Plot on number line
   -4… -3/2…. -2/5….. 4 [/su_spoiler]
3. I. x² – 652 = 504,
   II. y = √1156
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   x² – 652 = 504
   x² = 1156
   So x = 34, -34
   y = √1156 = 34
   Plot on number line
   -34… 34
   [/su_spoiler]
4. I. 9/√x + 8/(√x +1) = 5,
   II. 12/√y – 4/√y = 2
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   9/√x + 8/(√x +1) = 5
   [9(√x +1) + 8√x]/[√x * (√x +1)] = 5
   17√x + 9 = 5 (x + √x)
   5x – 12√x – 9 = 0
   5x – 15√x + 3√x – 9 = 0
   5√x (√x – 3) + 3 (√x – 3) = 0
   √x cannot be -3/3
   So √x = 3, so x = 9
   12/√y – 4/√y = 2
   8/√y = 2
   So √y = 4 or y = 16
   So y > x [/su_spoiler]
5. I. 3x² – 6x – √3x + 2√3 = 0,
   II. 2y² – 3y – 2√2y + 3√2 = 0,
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   3x² – 6x – √3x + 2√3 = 0
   3x (x- 2) – √3 (x – 2) = 0,
   So x = 2, √3/3
   2y² – 3y – 2√2y + 3√2 = 0
   y (2y – 3) – √2 (2y – 3) = 0
   So y = 3/2, √2 (1.44)
   plot on number line
   √3/3(0.57)…….√2…..(3/2)……2 [/su_spoiler]
6. I. x² – 2x – √5x + 2√5 = 0
   II. y² – 3y – √6y + 3√6 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   x² – 2x – √5x + 2√5 = 0
   x (x – 2) – √5 (x – 2) = 0
   So x = 2, √5 (2.23)
   y² – 3y – √6y + 3√6 = 0
   y (y – 3) – √6 (y – 3) = 0
   So y = 3, √6 (2.44)
   Plot on number line
   2…2.23……2.44…….3[/su_spoiler]
7. I. 8x² + 6x + 1 = 0,
   II. 5y² + 8y – 4 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relationship cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   8x² + 6x + 1 = 0
   8x² + 4x + 2x + 1 = 0
   So x = -1/4, -1/2
   5y² + 8y – 4 = 0
   5y² + 10y – 2y – 4 = 0
   So y = -2, 2/5
   [/su_spoiler]
8. I. 4x² – 23x + 30 = 0,
   II. 4y² – 3y – 45 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   4x² – 23x + 30 = 0
   4x² – 15x – 8x + 30 = 0
   So x = 15/4, 2
   4y² – 3y – 45 = 0
   4y² + 12y – 15y – 45 = 0
   So y = -3, 15/4
   Put on number line
   -3…. 2…. 15/4[/su_spoiler]
9. I. 5x² – 7x – 6 = 0,
   II. 3y² – 2y – 8 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   5x² – 7x – 6 = 0
   5x² – 10x + 3x – 6 = 0
   So x = -3/5, 2
   3y² – 2y – 8 = 0
   3y² – 6y + 4y – 8 = 0
   So y = -4/3, 1
   Plot on number line
   -4/3……-3/5….. 1….. 2[/su_spoiler]
10. I. 3x² + 2x – 21 = 0,
    II. 3y² – 19y + 28 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
    Solution:
    3x² + 2x – 21 = 0
    3x² + 9x – 7x – 21 = 0
    Gives x = -3, 7/3
    3y² – 19y + 28 = 0
    3y² – 12y – 7y + 28 = 0
    So y = 7/3, 4
    Put on number line
    -3……7/3……4 [/su_spoiler]

Directions (1 – 5): Study the following table carefully and answer the questions that follow:
A family buys different types of fruits – A, B, C, D each day. The following data shows the quantity and price of these fruits from Monday to Friday in a given week.
Q = Quantity in kg. P = Price/kg.

1. Which type of fruit did the family buy the maximum in the given period?
   A) A
   B) B
   C) C
   D) D
   E) Both D and C
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Fruit A bought = 2+4+3+3.5+4.5+6+5= 28 kg
   Fruit B bought = 30 kg
   Fruit C bought = 31 kg
   Fruit D bought = 31.5 kg [/su_spoiler]
2. On which of the following days did the family spent less on buying fruits?
   A) Thursday
   B) Friday
   C) Tuesday
   D) Monday
   E) Wednesday
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   Find only for days given in options
   For Thursday = (24*3.5 + 18*4.5 + 24*4.5 + 25*2.5) = 335.5
   For Friday = 322
   For Tuesday = 304
   For Monday = 290
   For Wednesday = 251 [/su_spoiler]
3. Total expenses on Sunday for buying fruits A and C is what percent of the total expenses on Saturday for buying fruits B and D?
   A) 126.28%
   B) 131.27%
   C) 178.45%
   D) 107.34%
   E) 123.78%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Expenses on Sunday for buying fruits A and C = 278
   Expenses on Saturday for buying fruits B and D = 259
   So required % = 278/259 * 100 = 107.34% [/su_spoiler]
4. For which of the following fruits, amount spent on Wednesday increases/decrease by more than 15% with respect to the previous day?
   A) Both A and C
   B) Both B and D
   C) Only A
   D) Both A and D
   E) Only C
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   For A amount spent, On Wed = 57, On Tuesday = 74, % decrease = 17/74 * 100 = 22.9%
   For B amount spent, On Wed = 74, On Tuesday = 77, % decrease = 3/77 * 100 = 3.9%
   For C amount spent, On Wed = 54, On Tuesday = 63, % decrease = 9/63 * 100 = 14.3%
   For D amount spent, On Wed = 66, On Tuesday = 90, % decrease = 24/90 * 100 = 26.67%
   [/su_spoiler]
5. Find the ratio between the total amounts spent on costliest fruit to cheapest fruit on Sunday.
   A) 22 : 13
   B) 23 : 15
   C) 20 : 9
   D) 21 : 11
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   On Sunday,
   Amount spent on A = 110
   Amount spent on B = 102
   Amount spent on C = 168
   Amount spent on D = 88
   So required ratio = 168 : 88 = 21 : 11 [/su_spoiler]

Directions (6-10): Study the following pie-chart to answer the questions that follow:

A shopkeeper has five products – A, B, C, D and E in his shop. The sale of products in a week is shown in the pie-chart. Given that the product A is sold at Rs 55 each and the shopkeeper makes Rs 58,850 by selling products A during the week.

6. What is the number of A products sold in the week?
   A) 1130
   B) 890
   C) 780
   D) 1070
   E) 990
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   58,850/55 = 1070 [/su_spoiler]
7. If on each D product, a profit of Rs 22 is made, find how much profit the shopkeeper made by selling products D in the week?
   A) Rs 13400
   B) Rs 18700
   C) Rs 12400
   D) Rs 16500
   E) Rs 14900
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   A products sold = 1070
   Let total number of all products sold = x
   So 107/360 * x = 1070, Solve x = 3600
   D products sold = 75/360 * 3600 = 750
   So profit on these 750 D products = 750*22 = Rs 16500 [/su_spoiler]
8. If only 84% of the total available C products in the shop were sold, then what is the number of unsold C products?
   A) 110
   B) 140
   C) 120
   D) 150
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   Total products sold = 3600
   So C products sold = 63/360 * 3600 = 630
   Now 84% were sold. So 84/100 * x = 630
   So total C products in the shop were x = 750
   So unsold C products = 750-630 = 120[/su_spoiler]
9. What is the ratio between the number of sold A and E products together and remaining products together?
   A) 7 : 11
   B) 8 : 13
   C) 5 : 11
   D) 8 : 15
   E) 9 : 15
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   (107+33) : (82+63+75) = 7 : 11 [/su_spoiler]
10. If cost price of each B product was Rs 250 and the shopkeeper earned a profit of 14% on each B product sold, find the total amount he got by selling B product?
    A) Rs 2,45,400
    B) Rs 2,33,700
    C) Rs 2,98,600
    D) Rs 2,28,800
    E) Rs 2,49,600
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
    Solution:
    B products sold are = 82/360 * 3600 = 820
    He sold 1 product at = 114/100 * 250 = Rs 285
    So total amount he got by selling B product = 285 * 820 = Rs 2,33,700 [/su_spoiler]

1. The area of the rectangle gets reduced by 10 if its length is increased by 2 and breadth decreased by 2. Also it increases by 14, if its breadth is increased by 3 and length decreased by 2. Find the original area of the rectangle.
   A) 126 sq. units
   B) 154 sq. units
   C) 173 sq. units
   D) 187 sq. units
   E) 163 sq. units
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   Let length = x, and breadth = y
   So xy – (x+2)(y-2) = 10
   Gives 2x – 2y + 4 = 10………..(1)
   And (x-2)(y+3) – xy = 14
   Gives 3x – 2y – 6 = 14………………(2)
   Solve equations, x = 14, y = 11 [/su_spoiler]
2. Incomes of Karan, Arjun and Aarti are in ratio 3 : 5 : 7. Their expenditures are in ratio 3 : 6 : 8. If B saves 10% of his income, find the ratio of their savings.
   A) 2 : 4: 5
   B) 5 : 4 : 2
   C) 4 : 3 : 2
   D) 3 : 2 : 4
   E) 3 : 4 : 2
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   Incomes = 3x, 5x, 7x
   Expenditures = 3y, 6y, 8y
   B’s savings = 10/100 * 5x = x/2
   So x/2 = 5x – 6y
   Solve, x = 4y/3
   Now savings ratio is 3x – 3y : 5x – 6y : 7x – 8y
   Put x = 4y/3
   So ratio becomes 1 : 2/3 : 4/3 = 3 : 2 : 4
   [/su_spoiler]
3. Karan can complete a work in 60 days and Arjun can complete the same work in 5 days more than the number of days in which they both can complete work together. They both start the work and after 5 days, Karan leaves and Arjun starts working with 4/3 times efficiency as before. In how much time the work will be completed?
   A) 12
   B) 8
   C) 10
   D) 15
   E) 18
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   Let they both complete work together in x days
   So Arjun in (x+5) days
   So 1/60 + 1/(x+5) = 1/x
   Solve, x = 15, So Arjun can complete work in 20 days
   Now they work for 5 days so completed (1/60 + 1/20) * 5 = 1/3 of work
   Remaining work = 2/3.  Arjun works with 4/3 times efficiency. So now he can complete whole work in 20/(4/3) = 15 days
   So 1/15 * y = 2/3
   Solve, y = 10
   So total number of days = 5 + 10 = 15[/su_spoiler]
4. Kavya borrowed Rs 2460 to be paid back in 2 yearly installments at 5% rate of interest compounded annually. How much will be each installment?
   A) Rs 1323
   B) Rs 1486
   C) Rs 1545
   D) Rs 1139
   E) Rs 1376
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   (1 + 5/100) = 21/20
   So for 2 yearly installments
   20x/21 + 400x/441 = 2460
   Solve, x = Rs 1323  [/su_spoiler]
5. A cylindrical bucket whose diameter is 14 m is partly filled with water. A cuboid whose dimensions are 33cm × 14cm × 7cm is immersed in the bucket. Find the level of water that rise?
   A) 22 cm
   B) 15 cm
   C) 18 cm
   D) 14 cm
   E) 21 cm
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   Volume of water that rise = Volume of cuboid
   So πr²h = l*b*h
   22/7 * 7 * 7 * h = 33 * 14 * 7
   Solve, h = 21 cm [/su_spoiler]
6. A loss of 10% is made by selling an article. Had it been sold for Rs 75 more, there would have been a profit of 5%. The initial loss is what percentage of the profit earned, if the article was sold at a profit of 20%?
   A) 44%
   B) 47%
   C) 50%
   D) 54%
   E) Cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   Use shortcut for these type of questions:
   CP of article = 75 × 100/ [5 – (-10)] = Rs 500  (+5 for 5% profit, -10 for 10% loss)
   So loss was = 10/100 * 500 = Rs 50
   and new profit = 20/100 * 500 = Rs 100
   So required % = 50/100 * 100 = 50% [/su_spoiler]
7. Vanya inquires about a mobile at two shops. One offers 30% discount and the other offers 25%. She has just sufficient amount of Rs 21,000 to purchase mobile at 30% discount, how much amount she has less than the amount required to purchase mobile at 25% discount?
   A) Rs 1600
   B) Rs 1500
   C) Rs 1700
   D) Rs 1300
   E) Rs 1400
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
   Solution:
   Let MP of mobile = Rs x
   So at 30% discount, she gets mobile at Rs 70% of x
   So 7x/10 = 21000
   Solve, x = 30,000
   So SP at 25% discount = 75% of 30,000= Rs 22,500
   Required difference =  22500 – 21000 = Rs 1500[/su_spoiler]
8. Two containers contain mixture of milk and water. Container A contains 25% water and rest milk. Container B contains 36% water and rest milk. How much amount should be mixed from container A to 50 litres of container B so as to get a new mixture having water to milk in ratio 2 : 5?
   A) 162 l
   B) 104 l
   C) 135 l
   D) 178 l
   E) 150 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option B
   Solution:
   In resultant mixture, water is 2/7 * 100 = 200/7%
   So by method of allegation:
   25%……………………………36%
   …………..200/7%
   52/2%……………………….25/2%
   So ratio is 52/2 : 25/2 = 52 : 25.
   So 52/25 = x/50
   So x = 104 l [/su_spoiler]
9. A boat travels downstream from point A to B and comes back to point C which is half distance between points A and B in a total of 40 hours. If the speed of boat is still water is 7 km/hr and distance from point A to B is 120 km, then find the upstream speed.
   A) 5 km/hr
   B) 8 km/hr
   C) 2 km/hr
   D) 3 km/hr
   E) 6 km/r
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  <strong> Option C
    Solution: </strong>
   A to B is 120, so B to is 120/2 = 60 km
   Let speed of current = x km/hr
   So 120/(7+x) + 60/(7-x) = 40
   Solve, x = 5 km/hr
   So downstream speed = 7 – 5 = 2 km/hr [/su_spoiler]
10. Ratio of age of A 2 years hence to age of C 5 years ago is 12 : 5. The average of ages of A and B is 20 years. Also age of B 6 years hence will be eight-five times of the present age of C. Find the age of A.
    A) 16 years
    B) 22 years
    C) 29 years
    D) 34 years
    E) 38 years
    [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
    Solution:
    (A+2)/(C-5) = 12/5
    (A+B)/2 = 20
    (B + 6) = 8/5 * C [/su_spoiler]

Directions: What will come in place of (?) in following series following a certain pattern?

1. 2, 7, 29, 109, 319, ?
   A) 724
   B) 704
   C) 629
   D) 835
   E) 853
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Pattern:
   2 × 6 – 5 = 7,
   7 × 5 – 6 = 29,
   29 × 4 – 7 = 109, 
   109 × 3 – 8 = 319, 
   319 × 2 – 9 = 629 [/su_spoiler]
2. 729, 243, 162, 162, 216, ?
   A) 555
   B) 456
   C) 782
   D) 360
   E) 685
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Pattern:
    × 1/3, × 2/3, × 3/3, × 4/3, × 5/3
   [/su_spoiler]
3. 216, 95, 176, 127, 152, ?
   A) 112
   B) 101
   C) 143
   D) 120
   E) 132
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Pattern:
   -11², + 9², -7², +5², -3² [/su_spoiler]
4. 3, 3, 7, 25, 73, ?
   A) 155
   B) 173
   C) 135
   D) 196
   E) 185
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Pattern:
   3 + (1² * 0) = 3
   3 + (2² * 1) = 7
   7 + (3² * 2) = 25
   25 + (4² * 3) = 73  
   73 + (5² * 4) = 173   [/su_spoiler]
5. 16, 20, 39, 83, ?, 286
   A) 195
   B) 190
   C) 140
   D) 158
   E) 162
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Pattern:
   16………. 20……….. 39…………. 83………. 162……….. 286
   .    +4          +19           +44            +79          +124       
   .           +15          +25           +35          +45 [/su_spoiler]
6. 16, 20, 34, 28, 52, ?
   A) 36
   B) 45
   C) 67
   D) 84
   E) 72
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Pattern:
   16………. 20……….. 34…………. 28………. 52……….. 36
   .    +4           +14           -6              +24             -16       
   .           +10          -20          +30             -40 [/su_spoiler]
7. 26, 20, 37, 29, 48, 38, ?
   A) 62
   B) 59
   C) 71
   D) 54
   E) 75
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Pattern:
   Two series
   26 + 11 = 37, 37+11 = 48, 48+11 = 59
   20+9 = 29, 29+9 = 38 [/su_spoiler]
8. 33, 37, 28, 44, 19, ?
   A) 66
   B) 59
   C) 34
   D) 42
   E) 55
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Pattern:
   +2², -3², +4², -5², +6²
   [/su_spoiler]
9. 8, 20, 56, 164, 488, ?
   A) 1387
   B) 1345
   C) 1460
   D) 1430
   E) 1428
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Pattern:
   *3 – 4, *3 – 4, *3 – 4,  *3 – 4,  *3 – 4  [/su_spoiler]
10. 4, 10, 14, 48, 186, ?
    A) 570
    B) 688
    C) 753
    D) 862
    E) 936
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
    Pattern:
    4 × 1 + 6 = 10,
    10 × 2 – 6 = 14,  
    14 × 3 + 6 = 48,  
    48 × 4 – 6 = 186,  
    186 × 5 + 6 = 936    [/su_spoiler]

Directions (1 – 5): Study the following bar graph carefully and answer the questions that follow:
The bar graph shows the number of students who joined the school in a particular year and who left the school in same year. The number of students in 2010 = 2500

1. What is the percentage increase/decrease in the strength of the students in school from 2012 to 2013?
   A) 3.45% increase
   B) 5.67% decrease
   C) 4.08% increase
   D) 3.18% decrease
   E) Cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   In 2012, students are = 2500 + (300-250) + (250-350) = 2450
   And in 2013, students are = 2450 + (500-400) = 2550
   So % increase = (2550-2450)/2450 * 100 = 4.08% [/su_spoiler]
2. If in 2015, the school organized an annual function, in which 45% students took part and 30% of them won various prizes, then what % of students in 2015 won the prizes?
   A) 12.8%
   B) 16.6%
   C) 12.2%
   D) 13.5%
   E) 15.7%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   In 2015, students are = 2550 + (450-300) + (400-500) = 2600
   Students who took part in function = 45% of 2600
   And who won prizes = 30% of 45% of 2600 = 351
   So required % = 351/2600 * 100 = 13.5% [/su_spoiler]
3. If the given data is just an estimation, and in real the number of students who join the school in 2013 is 50 more, in 2015 is 70 more and in 2016 is 60 less, then what is the total number of students in school in years 2016? (Other values are same)
   A) 2830
   B) 2710
   C) 2820
   D) 2750
   E) 2700
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Students in 2012 = 2450,
   So In 2013 now, 2450 + (500+50 – 400) = 2600
   Now in 2014, number of students = 2600 + (450-300)= 2750 [There is no change in data in 2014]
   In 2015, = 2750 + (400+70 -500) = 2720
   In 2016, = 2720 + (550-60-500) = 2710 [/su_spoiler]
4. For which of the following years, the percentage increase/decrease is the highest with respect to the previous year?
   A) 2011
   B) 2014
   C) 2013
   D) 2016
   E) 2012
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   In 2011 = 50/2500 * 100 = 2%,
   In 2012 = 100/2550 * 100 = 3.9%,
   In 2013 = 100/2450 * 100 = 4.08%,
   In 2014 = 150/2550 * 100 = 5.88%,
   In 2016 = 50/2600 * 100 = 1.9%[/su_spoiler]
5. The number of students in years 2012 and 2016 together is what percent of the number of students in years 2011 and 2014 together?
   A) 81.34%
   B) 98.18%
   C) 105.23%
   D) 97.14%
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   In 2012 + 2016 = 2450 + 2650 = 5100
   In 2011 + 2013 = 2550 + 2700 = 5250
   Required % = 5100/5250 * 100 = 97.14% [/su_spoiler]

Directions (6-10): Study the following table to answer the questions that follow:
The table shows the number of products produced and sold by a company in respective years

6. What is the percentage of unsold C products in tears 2013 and 2015 together?
   A) 9.82%
   B) 8.23%
   C) 7.65%
   D) 8.92%
   E) 9.28%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Unsold C in 2013 = 300 – 283 = 17, in 2015 = 350-309 = 41
   So required % = (17+41)/(300+350) *100 = 8.92%
   [/su_spoiler]
7. What is the average number of all products sold in all the years together?
   A) 934
   B) 953
   C) 919
   D) 902
   E) 968
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   Product A sold in all years = 224 + 146 + 275 + 319 = 964
   Product B sold in all years = 808
   Product C sold in all years = 1155
   Product D sold in all years = 1004
   Product E sold in all years = 834
   So required average = (964+808+1155+1004+834)/5 = 953[/su_spoiler]
8. Number of unsold A and B products in 2013 and 2016 forms what percent of sold C and E products in 2015?
   A) 24.57%
   B) 22.38%
   C) 26.54%
   D) 20.39%
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   Unsold A and B in 2013 and 2016 = (250-224) + (180-152) + (350-319) + (280-262) = 103
   Sold C and E in 2015 = 309 + 196 = 505
   So required % = 103/505 * 100 = 20.39% [/su_spoiler]
9. The company was able to sold how much percent of D products in the given period?
   A) 80.32%
   B) 84.45%
   C) 85.63%
   D) 84.97%
   E) 81.61%
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   In all years, D products sold are = 237 + 273 + 180 + 314 = 1004
   And produced = 280 + 300+ 310 + 360 = 1250
   So required % = 1004/1250 * 100 = 80.32% [/su_spoiler]
10. What is the percentage of sold B and E products in 2014?
    A) 87.34%
    B) 84.23%
    C) 82.67%
    D) 74.66%
    E) 72.29%
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
    Solution:
    Sold B and E in 2014 = 166 + 206 = 372
    Total produced B and E in 2014 = 200+250 = 450
    Required % = 372/450 * 100 = 82.67% [/su_spoiler]

Quantitative Aptitude Questions for SBI PO, NIACL, NICL, IBPS PO, IBPS RRB and other exams

1. Pipe A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened to fill a tank. The time when the tank should be full, it was found that a leak pipe was also opened along with pipes A and B. It is turned off immediately and then after 5 minutes, the tank gets full. Find the time in which the leak pipe can empty the full tank?
   A) 720/49 minutes
   B) 728/49 minutes
   C) 829/49 minutes
   D) 832/49 minutes
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   Pipes A and B can fill the tank in – 15*20/(15+20) = 60/7 minutes
   But along with leak pipe, it took 5 minutes more. Let leak pipe can empty tank in x minutes
   So A and B were opened for (60/7) + 5 = 95/7 minutes
   And the leak was opened for 60/7 minutes
   So (7/60) * 95/7 – (1/x) * 60/7 = 1
   Solve, x = 720/49 minutes
   [/su_spoiler]
2. Two mixtures A and B contain milk and water in ratio 7 : 1 and 6 : 1 respectively. 48 litres of mixture A is mixed with some quantity of mixture B and the resultant mixture C has milk and water in ratio 13 : 2. If 14 litres of mixture C is taken out, find the remaining quantity of mixture C.
   A) 79 l
   B) 72 l
   C) 75 l
   D) 76 l
   E) 71 l
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   Milk in A = 7/8, in B = 6/7 and in C is 13/15
   So by allegation method
   (48 l)…………………………(x l)
   7/8……………………………6/7
   …………….13/15
   1/105……………………….1/120
   So ratio is 1/105 : 1/120 = 8 : 7
   So, 48/x = 8/7
   x = 42 l
   So quantity of C = 48+42 = 90l, SO after 14 l is removed, remaining = 90 – 14 = 76 l
   [/su_spoiler]
3. A person wants to sell his chair and table. If he sells them at a total of Rs 525, he gets a profit of 20% on table and 10% on chair. But if he sells them at a total of Rs 510, he gets a profit of 10% on table and 20% on chair. Find the cost price of chair.
   A) Rs 250
   B) Rs 200
   C) Rs 150
   D) Rs 320
   E) Rs 270
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   When the profit % gets reversed, following formula can be used.
   Cost price of (table + chair) = [(525+510)]/(110% + 120%) =  1035/(1.10 + 1.20) = 450
   And Cost price of (table – chair) = [(525 – 510)]/(120% – 110%) =  15/(1.20 – 1.10) = 50   
   Solve both equations, CP of chair = Rs 150[/su_spoiler]
4. A and B started a business. A invests Rs 400 less than B. A invests for 7 months while B invests for 8 months. If out of a total profit of Rs 7700, B got Rs 1100 more than A, then find the investment of B.
   A) Rs 2800
   B) Rs 2400
   C) Rs 2500
   D) Rs 2100
   E) Rs 2300
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option A
   Solution:
   A’s investment = Rs x so B’s investment = (x+400)
   So ratio of shares of A and B is
   x * 7 : (x+400) * 8
   Now according to question, (x+400) * 8/[15x + 3200] * 7700 = 1100 + 7x/[15x + 3200] * 7700
   So , (x+400) * 8/[15x + 3200] * 7700  – 7x/[15x + 3200] * 7700  = 1100
   Gives [8x + 3200 – 7x] /[15x + 3200] * 7700  = 1100
   Gives [x + 3200] /[15x + 3200]   = 1/7
   Solve, x = Rs 2400 [/su_spoiler]
5. The difference between Simple Interest and Compound Interest after 3 years on a certain sum of money is Rs 12.16. If the rate % is 4% per annum, find the sum.
   A) Rs 2700
   B) Rs 3300
   C) Rs 3600
   D) Rs 2500
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option D
   Solution:
   Formula for difference between SI and CI for 3 years
   Difference = P * r²/100² * (300+r)/100
   So 12.16 = P * 16/10000 * 304/100
   So P = 1216 * 10000/(16 * 304) = 76*100*100/304 = Rs 2500
    [/su_spoiler]
6. Ratio between the ages of Shilpa and Meera is 4 : 5 and that between Shilpa and Sheetal is 2 : 3. If the sum of ages and Shilpa and Meera is one and a half times of the age of Sheetal. Find the total sum of ages of all three.
   A) 64
   B) 87
   C) 83
   D) 76
   E) Cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
   Solution:
   Meera/Shilpa = 5/4 and Shilpa/Sheetal = 2/3
   So Meera : Shilpa : Sheetal = 5*2 : 4*2 : 4*3 = 5 : 4 : 6
   Also given = (Shilpa + Meera) = 3/2 * Sheetal
   So (4x + 5x) = 3/2 * (6x) gives 9x = 9x, so cant be determined  [/su_spoiler]
7. A bucket contains 2 red balls, 4 blue balls, and 6 white balls. Two balls are drawn at random. What is the probability that either both are blue or both are non-blue?
   A) 5/11
   B) 14/33
   C) 12/33
   D) 6/11
   E) 17/33
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option E
    Solution:
   Two cases
   Case 1: both are blue
   Prob = ⁴C₂ / ¹²C₂ = 3/33
   Case 2: both non-blue (means red or white)
   Prob = ⁸C₂  / ¹²C₂ = 14/33.
   Add both cases [/su_spoiler]
8. Amit, Aman and Anil start a business by investing Rs 2500, Rs 3500 and Rs x respectively. They invest for 7 months, 5 months and 7 months respectively.  If out of a total profit of Rs 6400, Anil got Rs 2400, find the total investment of all three.
   A) Rs 9150
   B) Rs 9600
   C) Rs 9000
   D) Rs 9250
   E) None of these
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option C
   Solution:
   Ratio of shares of Amit : Aman : Anil is
   2500*7 : 3500*5 : x*7
   = 2500 : 2500 : x
   So x/(5000+x) * 6400 = 2400
   Solve, x = Rs 3000
   So total investment = 2500 + 3500 + 3000 = Rs 9000
   [/su_spoiler]
9. A vessel is in the form of a hemi-spherical bowl on which is mounted a hollow cylinder. The diameter of the sphere is 14 cm and the total height of vessel is 25 cm, find the capacity of the vessel.
   A) 3450.45 cu cm
   B) 4450.28 cu cm
   C) 3490.67 cu cm
   D) 4950.49 cu cm
   E) 3830.27 cu cm
   [su_spoiler title=” View Answer ” icon=”chevron-circle”]   Option C
   Solution: 
   Diameter is 14, so radius is 7 cm
   Total height = 25 cm, so height of cylinder = 25-7 = 18 cm (because height of hemisphere is same as its radius)
   Capacity of vessel = volume of cylinder + vol of hemisphere
   So                         = πr²h + 2/3 *πr³
   .                            = 22/7 * 7 * 7 * 18 + 2/3 * 22/7 * 7 * 7 * 7
   .                            = 2772 + 718.67
   .                            = 3490.67 cu cm
    [/su_spoiler]
10. The average age of 8 persons increased by 2 when a new person comes in place of one of them whose age is 26 kg. What might be the age of the new person?
    A) 36 years
    B) 42 years
    C) 39 years
    D) 44 years
    E) 48 years
    [su_spoiler title=” View Answer ” icon=”chevron-circle”]  Option B
     Solution:
    Total age increased = (8 * 2) = 16 years.
    Age of new person = (26 + 16) = 42 years. [/su_spoiler]

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly –

1. I. 20x² – 31x + 12 = 0,
   II. 3y² – 5y + 2 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   20x² – 31x + 12 = 0
   20x² – 16x – 15x + 12 = 0
   So x = 3/4, 4/5
   3y² – 5y + 2 = 0
   3y² – 3y – 2y + 2 = 0
   So y = 1, 2/3
   [/su_spoiler]
2. I. 3x² – 19x + 30 = 0,
   II. 3y² – 10y + 3 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option C
   Solution:
   3x² – 19x + 30 = 0
   3x² – 9x – 10x + 30 = 0
   So x = 3, 10/3
   3y² – 10y + 3 = 0
   3y² – 9y – y + 3 = 0
   So y = 1/3, 3
   [/su_spoiler]
3. I. 3x² – 25x + 52 = 0,
   II. 5y² – 18y + 9 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option A
   Solution:
   3x² – 25x + 52 = 0
   3x² – 12x – 13x + 52 = 0
   So x = 4, 13/3
   5y² – 18y + 9 = 0
   5y² – 15y – 3y + 9 = 0
   So y = 3/5, 3
   Put on number line
   -1/2 …. 3/5 ….2/3 …. 3
   [/su_spoiler]
4. I. 4x² – 5x – 6 = 0,
   II. 5y² – 7y – 6 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   4x² – 5x – 6 = 0
   4x² – 8x + 3x – 6 = 0
   So x = -3/4, 2
   5y² – 7y – 6 = 0
   5y² – 10y + 3y – 6 = 0
   So y = -3/5, 2
   [/su_spoiler]
5. I. 3x² – 10x + 8 = 0,
   II. 3y² – 14y + 16 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   3x² – 10x + 8 = 0
   3x² – 6x – 4x + 8 = 0
   So x = 2, 4/3
   3y² – 14y + 16 = 0
   3y² – 6y – 8y + 16 = 0
   So y = 2, 8/3
   [/su_spoiler]
6. I. 2x² + 17x + 30 = 0,
   II. 4y² – 7y – 15 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   2x² + 17x + 30 = 0
   2x² + 12x + 5x + 30 = 0
   So x = -6, -5/2
   4y² – 7y – 15 = 0
   4y² – 12y + 5y – 15 = 0
   So y = -5/4, 3
   [/su_spoiler]
7. I. 3x² + 16x + 20 = 0,
   II. 5y² + 8y – 4 = 0
   A) x > y
   B) x < y
   C) x ≥ y
   D) x ≤ y
   E) x = y or relationship cannot be determined
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option D
   Solution:
   3x² + 16x + 20 = 0
   3x² + 6x + 10x + 20 = 0
   So x = -10/3, -2
   5y² + 8y – 4 = 0
   5y² + 10y – 2y – 4 = 0
   So y = -2, 2/5
   [/su_spoiler]
8. I. 2x² + 17x + 21 = 0,
   II. 2y² + 13y + 15 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
   Solution:
   2x² + 17x + 21 = 0
   2x² + 14x + 3x + 21 = 0
   So x = -7, -3/2
   2y² + 13y + 15 = 0
   2y² + 10y + 3y + 15 = 0
   So y = -5, -3/2
   [/su_spoiler]
9. I. 5x² – 7x – 6 = 0,
   II. 3y² – 19y + 28 = 0
   A) If x > y
   B) If x < y
   C) If x ≥ y
   D) If x ≤ y
   E) If x = y or relation cannot be established
   [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option B
   Solution:
   5x² – 7x – 6 = 0
   5x² – 10x + 3x – 6 = 0
   So x = -3/5, 2
   3y² – 19y + 28 = 0
   3y² – 12y – 7y + 28 = 0
   So y = 7/3, 4
   [/su_spoiler]
10. I. 8x² + 6x – 5 = 0,
    II. 2y² + 7y – 4 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    [su_spoiler title=” View Answer ” icon=”chevron-circle”] Option E
    Solution:
    8x² + 6x – 5 = 0
    8x² – 4x + 10x – 5 = 0
    So x = -5/4, 1/2
    2y² + 7y – 4 = 0
    2y² + 8y – y – 4 = 0
    So y = -4, 1/2
    [/su_spoiler]