## Quant Arithmetic Questions /Word-Problem Quiz

Hello Aspirants, As we all know that Quant Arithmetic Questions is an important part of the quantitative aptitude section for every competitive exam. Different kinds of Quant Arithmetic questions are asked in every competitive exam. (Like – Percentage, Profit & Loss, Time and Work, Ratios and Proportions, Permutations and Combinations, Time and Distance, Boat and Streams, Probability, Average, Simple Interest, Compound Interest, Problems on Ages,). So here, In this article, we will provide different arithmetic questions. These quant arithmetic questions are important for Bank, SSC, SEBI, NABARD, RBI, LIC, and Other state exams. You can attempt these questions & boost your preparation for your examination.

In the Banking exams, Quant questions were asked in the Prelims as well as Mains exam. There are 10 Quant Miscellaneous Questions asked in the Prelims exam (Bank). You want to score more in the Quant section then you should practice more and more Quant Miscellaneous questions.

## Quant Arithmetic Questions Quiz | Set 19

1. A merchant sells 3 articles. First at a profit of 20%, second at a loss of 10% and third at the loss of 20%. If their selling price is same and the loss incurred on the whole transaction is Rs. 140x, then cost price of the second article is?

A. Rs. 800x
B. Rs. 480x
C. Rs. 400x
D. Rs. 560x
E. None of these

Correct Answer – A. Rs. 800x

Explanation: 2. Pipe A can fill a tank in 4 hours while pipe B can fill it in 6 hours working separately. Pipe C can empty whole the tank in 4 hours. He opened the pipe A and B simultaneously to fill empty tan. He wanted to adjust his alarm so that he could open the pipe C when it was half- filled, but he mistakenly adjusted his alarm at a time when his tank would be 3/4th filled. what is the time different between both case to fill the tank fully?

A. 60 min
B. 54 min
C. 48 min
D. Can’t be determine
E. None of these

Correct Answer – B. 54 min

Explanation: time according to half at alarm= (6/5 + 6/2)= 4.2 hours
time according 3/4 at alarm= (9/5+ 3/2) = 3.3 hours
difference of time= 4.2-3.3= .9 hours

=.9*60
=54 min

3. Wine W is produced by mixing alcohol X and alcohol Y in the ratio of 4 : 5. Alcohol X is prepared by mixing two raw materials, A and B in the ratio of 1 : 2. Alcohol Y is prepared by mixing two raw materials, B and C, in the ratio of 3 : 1. Then the final mixture is prepared by mixing 864 litres of wine W with water. If the concentration of the raw material B in the final mixture is 50%, then the quantity of water in the final mixture is_____

A. 372 litres
B. 373 litres
C. 368 litres
D. 432 litres
E. None of these

Correct Answer – C. 368 litres

Explanation: 4. Vasu, Ajay and Tina alone can do a work in 40 days, 25 days and 50 days respectively. Vasu started the work alone and after x days Tina joined him and after (x + 4) days Ajay joined them. The work was completed in y days.

Which of the following can be the value of x and y respectively?

A) 26, 32

B) 12, 22

C) 5.6, 17.6

A. Only A and B
B. Only B and C
C. Only A and C
D. Only C
E. All A, B and C

Correct Answer – C. Only A and C

Explanation:  Let, total work = LCM of 40, 25 and 50 = 200 units

Efficiencies of Vasu, Ajay and Tina is 5 units/day, 8 units/day and 4 units/day respectively.

Units of work done by Vasu and Tina in 4 days = 9 × 4 = 36 units

Remaining work = 200 – 36 = 164 units

Option A. 26, 32

So, 5 × 26 + 36 + 17 × 2 = 130 + 36 + 34 = 200 units

Option A is correct.

Option B. 12, 22

So, 5 × 12 + 36 + 17 × 6 = 198 units

Option B is not correct.

Option C. 5.6, 17.6

So, 5 × 5.6 + 36 + 17 × 8 = 28 + 36 + 136 = 200 units

Option C is correct.

5. In two colleges XAVIER and JOSEPH, the number of boys is more than the number of girls (Only boys and girls study in these two colleges). The ratio between the number of boys in XAVIER college to the difference in the number of boys and girls in JOSEPH college is 4 : 3. If the number of girls in JOSEPH college is 25% less than that in XAVIER college and the total number of students in XAVIER college is 200, then the number of boys in JOSEPH college?

A. 150
B. 175
C. 165
D. 140
E. None of these

Explanation:  Let the number of girls in XAVIER college be 4x, and number of boys in XAVIER college be 4y, then

Number of girls in JOSEPH college = 4x – 4x × 25% = 4x – x = 3x

 XAVIER college JOSEPH college Boys Girls Boys Girls 4y 4x 3x

According to question

4y : (Number of boys in JOSEPH college – 3x) = 4 : 3

⇒ Number of boys in JOSEPH college = 3x + 3y = 3(x + y)

Also given, 4y + 4x = 200

⇒ x + y = 50

Hence, the number of boys in JOSEPH college = 3(x + y) = 3 × 50 = 150

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6. A’s investment is half of the initial investment of B. B withdraws his money after 4 months. C joins the business after B left but not in the same month. C joins with amount X. If the shares of A and C in annual profit are same and B’s initial investment was Rs. 2400, then the which of the following is(are) the possible value of X.  (Assume that everyone invested for integer number of months)
I. Rs. 1800
II. Rs. 7200
III. Rs. 3600
IV. Rs. 1200
V. Rs. 5400

A. I and II only
B. II and III only
C. I, II and V only
D. II, IV and V only
E. None of these

Correct Answer – B. II and III only

Explanation: B’s initial investment = Rs. 2400, then
A’s initial investment = B’s initial investment/2 = 2400/2 = Rs. 1200
Let C invest for a months, then
The value of a must be integer less than 8 (As C joins the business after B left but not in the same month)

Ratio of equivalent investments of A, B and C = (1200 × 12) : (2400 × 4) : (X × a) = 14400 : 9600 : Xa
According to question
Xa = 14400
⇒ a = 14400/X
I. a = 14400/X = 14400/1800 = 8 (not possible)
II. a = 14400/X = 14400/7200 = 2 (possible)
III. a = 14400/X = 14400/3600 = 4 (possible)
IV. a = 14400/X = 14400/1200 = 12 (not possible)
V. a = 14400/X = 14400/5400 = 2.67 (not possible)
So, only II and III are true.

7. Area of a circle is 154 cm2. Width of a rectangle is 6 cm more than the radius of the circle. Length of rectangle is 3 cm more than its width. Find the difference between area of rectangle & area of circle.

A. 48 cm2
B. 54 cm2
C. 36 cm2
D. 64 cm2
E. None of these

Correct Answer – B. 54 cm2

Explanation: 22/7  × r2 = 154

r= 49

r = 7 cm

Width of rectangle = 6 + 7 = 13 cm

Length of rectangle = 13 + 3 = 16 cm

Area of rectangle = 16 × 13 = 208 cm2

Difference between the area of rectangle and circle = 208 – 154

= 54 cm2

8. A invested 50% of the amount invested by B . B withdrew the amount after 4 months. C joins the business after B leaves with an investment of Rs _______ for _______ months. At the end of the year A and C share the same amount of profit. The investment of B was Rs 3600. Which of the following values can fill the blanks?

I. 10800, 5

II. 3600, 6

III. 2700, 8

IV. 7200, 4

A. Only II and III
B. Only I and II
C. I, II, and IV
D. I, II, III, IV
E. None of these

Correct Answer – A. Only II and III

Explanation:

Let C invest amount x for a period of t months

A: B: C = 1800 × 12 : 3600 × 4 : x × t

The value of t for C can be maximum 8 months and minimum 1 month.

x × t = 1800 × 12

= 21600

Option I:

t = 5 x = 10800

x × t = 44000

Option II:

t = 6 x = 3600

x × t = 21600

Option III:

t = 8 x = 2700

x × t = 21600

Option IV:

t = 4 x = 7200

x × t = 28800

Only II and III satisfy

9. Speed of boat P in still water is 33 more than speed of boat Q in still water and speed of current is 16 km/hr. Boat Q starts from point X in downstream, 4 hours earlier than boat P.(both start from X downstream) It is found that boat P is 64 km ahead of boat Q after ‘t’ hours(t is an integer) from its start. If maximum speed of boat P in still water can be 64 km/hr and speed of boat Q in upstream will have integral value, which of the following can be the value of t?

I. 20 hours

II. 18 hours

III. 28 hours

IV. 32 hours

A. Only I
B. Only I and III
C. Only III, IV, and V
D. Only I, II and IV
E. Only I, II and III

Correct Answer – B. Only I and III

Explanation:

Let the speed of boat P and Q in still water be 4x km/hr and 3x km/hr respectively.

Using the data provided in the question, we get

(4x+16) – [(3x + 16) × (t + 4)] = 64

xt – 12x = 128

x × (t – 12) = 128

Max value of x = 16

t = 20 hrs

Min value of x = 8

t = 28 hrs

8 and 16 are the only possible values

t = 20 and 28

So, I and III can be the answer.

10. The ratio of the number of boys to the number of girls in a school of 720 students, is 7 : 5 . If 30 more boys are admitted in the school, then how many more girls should be admitted so that the ratio of boys to that of the girls, becomes 5 : 4 .

A. 60
B. 30
C. 45
D. 50
E. None of these

Explanation:

Number of boys = (7 / 12) × 720 = 420

Number of girls = (5 / 12) × 720 = 300

After admitted 30 more boys in the school the total number of boys

= 450,

Let admitted X number of girls in the school, then we can say

450 / 300+x = 5/4

1500 + 5x = 1800

5x = 300

x = 60 