**QUANT QUIZ FOR SSC EXAMS , Railways GROUP D & NTPC 2019**

**Q1. If a commission 0f 10% is given on mark price of Truck the dealer gains 20% if the commission is increase to 15% the gain % is?**

**(a)** 40/3

**(b)** 10

**(c)** 20

**(d)** 15

**(e)** None of these

**Q2. The CP of two dozen mangoes is Rs 32, after selling 18 mangoes at 12 Rs per dozen, the shopkeeper reduced the rate as Rs 4 per dozen. Then find the loss percentage?**

**(a)** 15

**(b)** 20

**(c)** 25

**(d)** 37.5

**(e)** None of these

**Q3. A can complete a task in 15 days B is 50% more efficient than A. Both A and B started working together on the task and after few days B left task and A finished the remaining 13 of the given work. For how many days A and B worked together.**

**(a)** 3

**(b)** 5

**(c)** 4

**(d)** 6

**(e)** 2

**Q4. A boat can travel 9.6 km downstream in 36 min. If speed of the water current is 10% of the speed of the boat in downstream, how much time will boat take to travel 19.2 km upstream?**

**(a)** 2 hours

**(b)** 3 hours

**(c)** 1.25 hours

**(d)** 1.5 hours

**(e)** 1 hour

**Q5. A started a business with a initial investment of Rs. 1200. ‘X’ month after the start of business, B joined A with on initial investment of Rs. 1500. If total profit was 1950 at the end of year and B’s share of profit was 750. Find ‘X’.**

**(a)** 5 month

**(b)** 6 month

**(c)** 7 month

**(d)** 8 month

**(e)** 9 month

** **

**Q6. Ratio between curved surface area and total surface area of a circular cylinder is 3: 5. If curved surface area is 1848 cm3 then what is the height of cylinder?**

**(a)** 28

**(b)** 14

**(c)** 17

**(d)** 21

**(e)** 7

**Q7. Train A completely crosses train B which is 205 m long in 16 second. If they are travelling in opposite direction and sum of speed of both are 25 m/s. then find the difference (in meter) between lengths of both trains.**

**(a)** 5

**(b)** 6

**(c) **8

**(d)** 10

**(e)** 12

**Q8. Present age of A is 3 years less than present age of B. Ratio of B’s age 5 year ago and A’s age 4 year hence is 3 : 4 then find present age (in years) of A.**

**(a)** 20

**(b)** 17

**(c)** 23

**(d)** 26

**(e)** 29

**Q9. Elena’s age after 15 years will be 5 times of her age 5 years back. What is the present age of Elena?**

**(a)** 10

**(b)** 37

**(c)** 35

**(d)** 11

**(e)** None of these

**Q10. A mess has provisions for 60 days. If after 15 days 500 men strengthen then and the food lasts 40 days longer, how many men are there in the mess?**

**(a) **3500

**(b) **4000

**(c) **6000

**(d) **8000

**(e) **None of these

**Directions (11-15): In each of the following questions two equations are given followed by 5 options. You have to solve the equations and give answer:**

** **

** **

**Q.11 I. 5x² + 29x + 20 = 0 **

**II. 25y² + 25y + 6 = 0 **

(a) if x < y

(b) if x ≤ y

(c) relationship between x and y cannot be determined

(d) if x ≥ y

(e) if x > y

** **

**Q.12 I. 3x² – 16x + 21 = 0**

**II. 3y² – 28y + 65 = 0 **

(a) if x < y

(b) if x ≤ y

(c) relationship between x and y cannot be determined

(d) if x ≥ y

(e) if x > y

** **

13 I. x² + x – 12 = 0

**II. y² + 2y – 8 = 0**

(a) if x < y

(b) if x ≤ y

(c) relationship between x and y cannot be determined

(d) if x ≥ y

(e) if x > y

** **

14 I. 4x² – 13x + 9 = 0

**II. 3y² – 14y + 16 = 0**

(a) if x < y

(b) if x ≤ y

(c) relationship between x and y cannot be determined

(d) if x ≥ y

(e) if x > y

**Q.15 I. 16x² + 20x + 6 = 0 **

**II. 10y² + 38y + 24 = 0 **

(a) if x < y

(b) if x ≤ y

(c) relationship between x and y cannot be determined

(d) if x ≥ y

(e) if x > y

**Solutions:**

**Sol 1.** (a) Let S.P. = Rs. 100

After commission, price = Rs. 90

CP = 100/120×90 = Rs. 75

Now, commission = 15%

Gain % = 85−75/75×100

** =40/3%**

**Sol 2.** (d) Total CP = Rs. 32

Total SP = 12 + 6 + 2 = Rs. 20

Loss % =12/32×100

**= 37.5%**

**Sol 3.** (c) A can complete work in 15 days.

And B will complete work in 10 days.

They together will complete whole work in =15×10/25=6 days

**A and B together worked for = 6 × 2/3= 4 days**

**Sol 4.** (d) Speed of downstream=9.6/36kmmin=16 km/hr

Speed of current = 1.6 km/hr

Let speed of man in still water = 𝑥

So, 𝑥 = 16 – 1.6 = 14.4 km/hr

Required time in upstream=19.2/14.4−1.6

** =1.5 hours**

**Sol 5.** (b) Ratio of profit of A and B = 1200:750 = 8:5

So, 1200×12/1500×𝑦 = 8/5

𝑦=6 months

** x= 6 month**

**Sol 6.** (d) 2πrh/2πr(r+h) = 3/5

2h = 3r

And 2πrh = 1848

2×22/7×2/3h×h=1848

** h=21**

**Sol 7.** (d) In 16 second distance covered by both A and B = 16 × 25 = 400 m

So length of A = 400 – 205 = 195

** Required difference = 10 m**

**Sol 8.** (a) Let B’s age = 𝑥

So A’s age = 𝑥−3

𝑥−5/𝑥+1=3/4

𝑥=23

**A’s age = 23−3=20 𝑦𝑒𝑎𝑟𝑠**

**Sol 9.** (a) Let Elena’s age = x

x + 15 = 5 (x – 5)

**x = 10 years**

**Sol 10.** (b) Let No. of soldiers = x

60 × x = 15x + 40 (x + 500)

60x =15x + 40x + 20000

5x = 20000

** x = 4000**

**S11. Ans. (a)****
**Sol.

I. 5x² + 29 + 20 = 0

⇒ 5x² + 25x + 4x + 20 = 0

⇒ (x + 5) (5x + 4) = 0

⇒ x = –5, –4/5

- 25y² + 25y + 6 = 0

⇒ 25y² + 15y + 10y + 6 = 0

⇒ (5y + 3) (5y + 2) = 0

⇒ y = – 3/5, –2/5

**y > x**

**S12. Ans.(a)**

Sol.

I. 3x² – 16x + 21 = 0

⇒ 3x² – 9x – 7x + 21 = 0

⇒ (x – 3) (3x – 7) = 0

⇒ x = 3, 7/3

II. 3y² – 28y + 65 = 0

⇒ 3y² – 15y – 13y + 65 = 0

⇒ (y – 5)(3y – 13) = 0

⇒ y = 5, 13/3

**y > x**

** **

**S13. Ans.(c)****
**Sol.

I. x² + x – 12 = 0

⇒ x² + 4x – 3x – 12 = 0

⇒ (x + 4) (x – 3) = 0

⇒ x = 3, –4

II. y² + 2y – 8 = 0

⇒ y² + 4y – 2y – 8 = 0

⇒ (y + 4) (y – 2) = 0

⇒ y = – 4, 2

**No relation**

**S14. Ans.(c)****
**Sol.

I. 4x² – 13x + 9 = 0

⇒ 4x² – 4x – 9x + 9 = 0

⇒ (x – 1) (4x – 9) = 0

⇒ x = 1, 9/4

II. 3y² – 14y + 16 = 0

⇒ 3y² – 6y – 8y + 16 = 0

⇒ (y – 2) (3y – 8) = 0

⇒ y = 2, 8/3

**No relation**

** **

**S15. Ans.(e)****
**Sol.

I. 16x² + 20x + 6 = 0

⇒ 8x² + 10x + 3 = 0

⇒ 8x² + 4x + 6x + 3 = 0

⇒ (2x + 1) (4x + 3) = 0

⇒ x = –1/2, –3/4

- 10y² + 38y + 24 = 0

⇒ 5y² + 19y + 12 = 0

⇒ 5y² + 15y + 4y + 12 = 0

⇒ (y + 3) (5y + 4) = 0

y = –3, –4/5

**x > y**

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