Quantitative Reasoning Questions for MAH-CET PDF

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Question 1: The rate of onion increased from ₹ 70 per kilogram to ₹ 120 per kilogram within a month. By what percentage (approximate) did the rate of onion increase?

a) 75.2%

b) 65%

c) 80%

d) 71.5%

1) Answer (D)

Solution:

Percentage increase in the rate of onion = $\frac{120-70}{70}\times100$

= $\frac{50}{70}\times100$

= $\frac{500}{7}$

= 71.43%

= 71.5% (approximately)

Hence, the correct answer is Option D

Question 2: Which two signs and two numbers should be interchanged to make the given equation correct?
$16 + 8 \times 25 – 15 \div 5 = 138$

a) $\times$ and $+$; 15 and 25

b) $\div$ and $-$; 15 and 5

c) $\times$ and $-$; 8 and 25

d) $\times$ and $+$; 8 and 25

2) Answer (A)

Solution:

By Trial and Error method,

Option A

$16\times8+15-25\div5=138$

$16\times8+15-5=138$

$128+15-5=138$

$138=138$

Hence, the correct answer is Option A

Question 3: Select the correct equation when the signs ‘+’ and $‘\times’$ and the numbers ‘4’ and ‘8’ are interchanged.

a) $12 \times 4 + 8 = 34$

b) $6 + 8 \times 4 = 38$

c) $2 \times 4 + 8 = 34$

d) $4 + 8 \times 2 = 32$

3) Answer (C)

Solution:

By Trial and Error method,

Option A

$12+8\times4=34$

$12+32=34$

$44=34$

Hence Option A is incorrect

Option B

$6\times4+8=38$

$24+8=38$

$32=38$

Hence Option B is incorrect

Option C

$2+8\times4=34$

$2+32=34$

$34=34$

Hence, the correct answer is Option C

Question 4: Train ‘A’ runs at a speed of 80 kmph and leaves station ‘X’ at 11.00 a.m. Train ‘B’ leaves station ‘X’ at 11.15 a.m.in the same direction, on the same day . At what speed train ‘B’ should run in order to catch train ‘A’ at station ‘Y’ located at a distance of 60 km?

a) 120 km/h

b) 125 km/h

c) 115 km/h

d) 110 km/h

4) Answer (A)

Solution:

Speed of train A = 80 km/h

Distance between station X and station Y = 60 km

Time taken by train A to reach station Y = $\frac{60}{80}$ hours = $\frac{3}{4}$ hours $\frac{3}{4}\times$60 minutes = 45 minutes

Since train B starts 15 min late, it should reach station Y in 30 minutes to catch train A at station Y.

Time taken by train B to reach station Y = 30 minutes = $\frac{1}{2}$ hour

$\therefore\ $Speed of train B = $\frac{\text{Distance from station X to station Y}}{\text{Time}}$ = $\frac{60}{\frac{1}{2}}$ = 120 km/h

Hence, the correct answer is Option A

Question 5: Which two numbers should be interchanged in the following equation to make it correct?
$36 \div 6 – 15 \times 2 + 48 = 14$

a) 36 and 48

b) 6 and 14

c) 48 and 15

d) 2 and 6

5) Answer (A)

Solution:

By Trial and Error method,

Option A

$48\div6-15\times2+36=14$

$8-15\times2+36=14$

$8-30+36=14$

$14=14$

Hence, the correct answer is Option A

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Question 6: In a class, every student participated in one of the three activities i.e. folk song, kathak and quiz. 22% of the total students participated in folk song. 25% of the remaining students participated in kathak. The remaining students participated in quiz. What percentage of students participated in quiz?

a) 53%

b) 30.6%

c) 41.5%

d) 58.5%

6) Answer (D)

Solution:

Let the total number of students = T

22% of the total students participated in folk song

Number of students participated in folk song = $\frac{22}{100}$T

Remaining students = T – $\frac{22}{100}$T = $\frac{78}{100}$T

25% of the remaining students participated in kathak

Number of students participated in kathak = $\frac{25}{100}\times\frac{78}{100}$T = $\frac{39}{200}$T

Remaining students participated in quiz.

Number of students participated in quiz = T – $\frac{22}{100}$T – $\frac{39}{200}$T = $\frac{117}{200}$T

Percentage of students participated in quiz = $\frac{\frac{117}{200}T}{T}\times100$ = 58.5%

Hence, the correct answer is Option D

Question 7: A person is employed to a company on salary of ₹ 6,000 per month. Through out the year, he remained absent from work for 15 days. If ₹ 200 gets deducted for every single holiday, then how much money will he earn at the end of the year ?

a) ₹ 72,000

b) ₹ 70,000

c) ₹ 69,000

d) ₹ 75,000

7) Answer (C)

Solution:

Salary of the person per month = ₹ 6,000

Salary of the person per year = $6000\times12$ = ₹ 72,000

Deduction for 1 holiday = ₹ 200

Deduction for 15 holidays = $15\times200$ = ₹ 3,000

$\therefore\ $Money earned by the person at the end of the year = 72000 – 3000 = ₹ 69,000

Hence, the correct answer is Option C

Question 8: What will be the value of ‘a + b’ in the following equation?
$\left[4 \left(3 + \left\{9 \times 6 \div 3 + \left\{12 \div \left(1 \div 1 \times 3\right)\right\}\right\}\right)\right] = 5^a \times 2^b$

a) 5

b) 6

c) 7

d) 4

8) Answer (D)

Solution:

$\left[4\left(3+\left\{9\times6\div3+\left\{12\div\left(1\div1\times3\right)\right\}\right\}\right)\right]=5^a\times2^b$

$\Rightarrow$  $\left[4\left(3+\left\{9\times6\div3+\left\{12\div\left(1\times3\right)\right\}\right\}\right)\right]=5^a\times2^b$

$\Rightarrow$  $\left[4\left(3+\left\{9\times6\div3+\left\{12\div3\right\}\right\}\right)\right]=5^a\times2^b$

$\Rightarrow$  $\left[4\left(3+\left\{9\times6\div3+4\right\}\right)\right]=5^a\times2^b$

$\Rightarrow$  $\left[4\left(3+\left\{9\times2+4\right\}\right)\right]=5^a\times2^b$

$\Rightarrow$  $\left[4\left(3+\left\{18+4\right\}\right)\right]=5^a\times2^b$

$\Rightarrow$  $\left[4\left(3+22\right)\right]=5^a\times2^b$

$\Rightarrow$  $\left[4\left(25\right)\right]=5^a\times2^b$

$\Rightarrow$  $100=5^a\times2^b$

$\Rightarrow$  $25\times4=5^a\times2^b$

$\Rightarrow$  $5^2\times2^2=5^a\times2^b$

Comparing both sides,  a = 2 and b = 2

$\therefore\ $a + b = 2 + 2 = 4

Hence, the correct answer is Option D

Question 9: One pound is approximately equal to 0.453 kg. Jennifer bought 4.409 pounds of custard apple. How many kilograms of custard apple did she buy (approximately)?

a) 2

b) 1

c) 1.5

d) 2.5

9) Answer (A)

Solution:

Given, One pound is approximately equal to 0.453 kg

Custard apple bought by Jennifer = 4.409 pounds = 4.409 x 0.453 kg = 1.997 kg = 2 kg (approximately)

Hence, the correct answer is Option A

Question 10: Select the correct combination of mathematical signs to replace ‘Y’ sequentially and balance the following equation.
18 Y 2 Y 3 Y 3 Y 9

a) $\times, =, -, \div$

b) $\times, \div, +, =$

c) $\div, \times, =, -$

d) $\div, \times, =, \times$

10) Answer (D)

Solution:

By Trial and Error method,

Option A

$18\times2=3-3\div9$

$18\times2=3-\frac{1}{3}$

$36=3-\frac{1}{3}$

$36=\frac{8}{3}$

Hence Option A is incorrect

Option B

$18\times2\div3+3=9$

$18\times\frac{2}{3}+3=9$

$12+3=9$

$15=9$

Hence Option B is incorrect

Option C

$18\div2\times3=3-9$

$9\times3=3-9$

$27=-6$

Hence Option C is incorrect

Option D

$18\div2\times3=3\times9$

$9\times3=3\times9$

$27=27$

Hence, the correct answer is Option D

Question 11: Which two signs and numbers should be interchanged to make the following equation correct?
$16 \times 18 + 2 – 14 \div 3 = 38$

a) 14 and 18, + and –

b) 16 and 14, – and $\times$

c) 14 and 18, + and $\times$

d) 16 and 3, – and $\div$

11) Answer (C)

Solution:

By Trial and Error method,

Option A

$16\times14-2+18\div3=38$

$16\times14-2+6=38$

$224-2+6=38$

$228=38$

Hence Option A is incorrect

Option B

$14-18+2\times16\div3=38$

$14-18+2\times\frac{16}{3}=38$

After solving the value will be decimal which is not possible

Option C

$16+14\times2-18\div3=38$

$16+14\times2-6=38$

$16+28-6=38$

$38=38$

Hence, the correct answer is Option C

Question 12: Which two signs should be interchanged to make the following equation correct?
$4 \div 6 + 9 – 48 \times 8 = 27$

a) + and $\times$

b) + and –

c) $\div$ and +

d) $\div$ and $\times$

12) Answer (D)

Solution:

By Trial and Error method,

Option A

$4\div6\times9-48+8=27$

$\frac{2}{3}\times9-48+8=27$

$6-48+8=27$

$-34=27$

Hence Option A is incorrect

Option B

$4\div6-9+48\times8=27$

$\frac{2}{3}-9+48\times8=27$

After solving the value will be decimal which is not possible

Option C

$4+6\div9-48\times8=27$

$4+\frac{2}{3}-48\times8=27$

After solving the value will be decimal which is not possible

Option D

$4\times6+9-48\div8=27$

$4\times6+9-6=27$

$24+9-6=27$

$27=27$

Hence, the correct answer is Option D

Question 13: A and B can do a piece of work in 30 days and 18 days respectively. A started the work alone and then after 6 days B joined him till the completion of the work. In how many days has the whole work completed?

a) 17

b) 15

c) 9

d) 12

13) Answer (B)

Solution:

Let the total work be W

Number of days required for A to complete the work = 30 days

$\Rightarrow$  Work done by A in 1 day = $\frac{W}{30}$

Number of days required for B to complete the work = 18 days

$\Rightarrow$  Work done by B in 1 day = $\frac{W}{18}$

Work done by A and B together in 1 day = $\frac{W}{30}+\frac{W}{18}$ = $\frac{3W+5W}{90}$ = $\frac{4W}{45}$

Work done by A alone in 6 days = $\frac{W}{30}\times6$ = $\frac{W}{5}$

Remaining work = $W-\frac{W}{5}$ = $\frac{4W}{5}$

Number of days required for both A and B to complete remaining work = $\frac{\text{Remaining work}}{\text{Work in 1 day}}$ = $\frac{\frac{4W}{5}}{\frac{4W}{45}}$ = 9 days

$\therefore\ $Number of days required to complete the whole work = 6 + 9 = 15 days

Hence, the correct answer is Option B

Question 14: Which two signs should be interchanged to make the given equation correct?
$(72 \div 18) + 30 \times 8 – 4 = 20$

a) + and –

b) + and $\times$

c) $\times$ and $\div$

d) + and $\div$

14) Answer (D)

Solution:

By Trial and Error method,

Option A

$(72\div18)-30\times8+4=20$

$4-30\times8+4=20$

$4-240+4=20$

$-232=20$

Hence Option A is incorrect

Option B

$(72\div18)\times30+8-4=20$

$4\times30+8-4=20$

$120+8-4=20$

$124=20$

Hence Option B is incorrect

Option C

$(72\times18)+30\div8-4=20$

$1296+30\div8-4=20$

$1296+3.75-4=20$

After solving the  value will be decimal which is not possible

Option D

$(72+18)\div30\times8-4=20$

$90\div30\times8-4=20$

$3\times8-4=20$

$24-4=20$

$20=20$

Hence, the correct answer is Option D

Question 15: Which two signs or numbers need to be interchanged to make the following equation correct?
$(18 \div 9) + 9 \times 8 = 24$

a) $+, \times$

b) $+, \div$

c) $\times, \div$

d) $18, 8$

15) Answer (B)

Solution:

By Trial and Error method,

Option A

$(18\div9)\times9+8=24$

$2\times9+8=24$

$18+8=24$

$26=24$

Hence Option A is incorrect

Option B

$(18+9)\div9\times8=24$

$27\div9\times8=24$

$3\times8=24$

$24=24$

Hence, the correct answer is Option B

Question 16: If ‘#’ means ‘—’, ‘&’ means ‘$\div$’, ‘@’ means ‘$\times$’ and ‘$\div$’ means ‘+’, then 15 @ 2 $\div$ 900 & 30 # 10 = ?

a) 21

b) 310

c) 50

d) 600

16) Answer (C)

Solution:

By applying the given conditions,

$15$ @ $2$ $\div$ $900$ & $30$ # $10$ $=15\times2\ +900 \div 30 -10$

$=15\times2\ +30\ -10$

$=30+30\ -10$

$=50$

Hence, the correct answer is Option C

Question 17: Which two numbers should be interchanged in the following equation to make it correct?
$6 + 28 \div 4 – 2 \times 17 = 12$

a) 17 and 4

b) 28 and 2

c) 17 and 6

d) 6 and 4

17) Answer (C)

Solution:

By Trial and Error method,

Option A

$6+28\div17-2\times4=12$

$6+\frac{28}{17}-2\times4=12$

After solving the value will be decimal which is not possible. Hence Option A is incorrect.

Option B

$6+2\div4-28\times17=12$

$6+0.5-28\times17=12$

After solving the value will be decimal which is not possible. Hence Option B is incorrect.

Option C

$17+28\div4-2\times6=12$

$17+7-2\times6=12$

$17+7-12=12$

$12=12$

Hence, the correct answer is Option C

Question 18: Aman has an equal number of one, five, ten and twenty rupee bank notes, to make a total of ₹ 1,080. How many bank notes does Aman have in total?

a) 100

b) 90

c) 120

d) 105

18) Answer (C)

Solution:

Let the number of one rupee notes with the Aman = a

Aman has an equal number of one, five, ten and twenty rupee bank notes

$\Rightarrow$  Total number of notes with Aman = 4a

Total amount with Aman = ₹ 1,080

$\Rightarrow$  $\left(1\times a\right)+\left(5\times a\right)+\left(10\times a\right)+\left(20\times a\right)=1080$

$\Rightarrow$  $36a=1080$

$\Rightarrow$  $a=30$

$\therefore\ $Total number of notes with Aman = 4a = 4 x 30 = 120

Hence, the correct answer is Option C

Question 19: Which two signs should be interchanged to make the given equation correct?
$(560\div80)+90\times8-38=600$

a) $+$ and $-$

b) $+$ and $\div$

c) $\div$ and $-$

d) $\times$ and $+$

19) Answer (D)

Solution:

By Trial and Error method,

Option A

$(560\div80)-90\times8+38=600$

$7-90\times8+38=600$

$7-720+38=600$

$-675=600$

Hence Option A is incorrect

Option B

$(560+80)\div90\times8-38=600$

$640\div90\times8-38=600$

$\frac{64}{9}\times8-38=600$

The value will be decimal after solving which is not possible

Option C

$(560-80)+90\times8\div38=600$

$480+90\times8\div38=600$

$480+720\div38=600$

$480+\frac{360}{19}=600$

The value will be decimal after solving which is not possible

Option D

$(560\div80)\times90+8-38=600$

$7\times90+8-38=600$

$630+8-38=600$

$600=600$

Hence, the correct answer is Option D

Question 20: After giving a discount of 25% on the marked price, a laptop is sold at ₹ 22,500. What is its marked price?

a) ₹ 32,000

b) ₹ 36,500

c) ₹ 30,000

d) ₹ 29,000

20) Answer (C)

Solution:

Let the marked price be M

Discount = 25%

Selling price = ₹ 22,500

$\Rightarrow$  $\frac{75}{100}\text{M}=22500$

$\Rightarrow$  $\text{M}=30000$

$\therefore\ $Marked price = ₹ 30,000

Hence, the correct answer is Option C

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