QUANT QUIZ (Misc.) FOR SSC CHSL EXAM

 



Q3. The ratio between two numbers is 3 : 4. If each number is increased by 6, the ratio becomes 4 : 5. The difference between the numbers is:
(a) 1
(b) 3
(c) 6
(d) 8
S3. Ans. (c)
Sol. Let the numbers be 3x and 4x. Then, 
(3x+6)/(4x+6)=4/5
⇒ 5(3x + 6) = 4(4x + 6)
⇒ x = (30 – 24) = 6

Difference between the numbers = (4x – 3x) = x = 6



Q5. Ram had to do a multiplication. Instead of taking 35 as one of the multipliers, he took 53. As a result, the product went up by 540. What is the new product? 
(a) 1050
(b) 1590
(c) 1440
(d) None of these 
S5. Ans.(b)
Sol. Suppose one of the multiplier is x, then 
53x – 35x = 540 ⇒ (18x = 540) 
∴ x = 30

∴ New multiplication = 30 × 53 = 1590 
Q6. Find the number nearest to 2559 which is exactly divisible by 35.   
(a) 2535
(b) 2555
(c) 2540
(d) 2550 



Q7. A man riding his bicycle covers 150 metres in 25 seconds. What is his speed in km per hour?  
(a) 20 km/hr 
(b) 21.6 km/hr 
(c) 23 km/hr 
(d) 25 km/hr
S7. Ans.(b)
Sol. Speed =150/25 m/sec =(150/25×18/5)km/hr 

=108/5 km/hr = 21.6 km/hr 

Q8. A takes 2 hours more than B to walk d km. If A doubles his speed then he can make it in 1 hour less than B. how much time does B require for walking d km?  
(a) d/2 hrs 
(b) 3 hrs 
(c) 4 hrs 
(d) 2d/3 hrs 
S8. Ans.(c)
Sol. Suppose B takes x hours to walk d km
Then, A takes (x + 2) hours to walk d km 
With Double of the speed, 
A will take 1/2 (x+2) hrs 
⇒ x-1/2 (x+2)=1 
⇒ 2x – (x + 2) = 2
⇒ x = 4 

Hence B takes 4 hours to walk d km. 


Q9. A train crosses a pole in 15 seconds, while it crosses a 100 m long platform in 25 seconds. The length of the train is:  
(a) 125 m
(b) 135 m
(c) 150 m
(d) 175 m
S9. Ans.(c)
Sol. Let the length of the train be x metres 
Then x/15=((100 + x))/25 ⇒ 25x = 1500 + 15x
⇒ 10x = 1500 ⇒ x = 150 

Hence, the length of the train is 150 m 


Q10. A began a business with Rs. 4500 and was joined afterwards by B with Rs. 5400. If the profits at the end of the year were divided in the ratio 2 : 1, B joined the business after:  
(a) 4
(b) 5
(c) 6
(d) 7
S10. Ans.(d)
Sol. Suppose B remained in the business for x months. 
Then, A : B =(4500×12)/(5400×x)=2/1 
⇒ 2x = 10 ⇒ x = 5

Thus, B remained in the business for 5 months. So, B joined the business after 7 months. 


Q11. Shruti took a loan at simple interest at 6% in the first year with an increase of 0.5% in each subsequent year. She paid Rs. 3375 as interest after 4 years. How much loan did she take?  
(a) Rs. 12500
(b) Rs. 15800
(c) Rs. 33250
(d) Cannot be determined 
S11. Ans.(a)
Sol. Let the loan taken be Rs. x 
Then,  x×6/100×1+x×6.5/100×1+x×7/100×1+x×7.5/100×1=3375 
⇒ (6+6.5+7+7.5)×x/100=3375 

⇒ x=(3375×100/27)=12500


Q12. At what rate of simple interest a certain sum will be double in 15 years?
(a) 5 (1/2) p.a. 
(b) 6% p.a. 
(c) 6 (2/3)% p.a. 
(d) 7 (1/2)% p.a. 
S12. Ans. (c)
Sol. Let the sum be Rs. x
Then, S.I. = Rs. x 
Rate =((100 × S.I.)/(Sum × Time))=((100 × x)/(x × 15))% p.a. 

=20/3%=6 (2/3)% p.a. 


Q13. The difference between compound interest and simple interest on a sum for 2 years at 8% p.a. is Rs. 768. The sum is  
(a) Rs. 100000
(b) Rs. 110000
(c) Rs. 120000
(d) Rs. 170000  

Q14. The perimeter of the triangular base of a right prism is 15 cm and radius of the 
incircle of the triangular base is 3 cm. If the volume of the prism be 270 cm3, then the height of the prism is  
(a) 6 cm  
(b) 7.5 cm 
(c) 10 cm 
(d) 12 cm 

Q15. The base of a right pyramid is a square of side 40 cm long. If the volume of the pyramid is 8000 cm3, then its height is:  
(a) 5 cm 
(b) 10 cm 
(c) 15 cm 
(d) 20 cm 
S15. Ans.(c)
Sol. Area of the base = 40 × 40 = 1600 sq. cm 
Volume of pyramid =1/3 × area of base × height 
⇒ 8000=1/3×1600×h

⇒ h=(8000 × 3)/1600=15 cm  

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