Solve these sums in a jiffy

Hyderabad: This article is in continuation to the last article focusing on the ratio and proportion topic. Here are some practice questions along with solutions on the ratio and proportion topic that will help you in your preparation for the State government recruitment jobs. The weight of Mr.Gupta and Mrs. Gupta are in the ratio […]

Hyderabad: This article is in continuation to the last article focusing on the ratio and proportion topic. Here are some practice questions along with solutions on the ratio and proportion topic that will help you in your preparation for the State government recruitment jobs.

The weight of Mr.Gupta and Mrs. Gupta are in the ratio 7 : 8 and their total weight is 120 kg. After taking a dieting course Mr.Gupta reduces by 6 kg and the ratio between their weights changed to 5 : 6, So, Mrs. Gupta has reduced by?
a) 2 kg b) 3 kg c) 4 kg d) 5 kg

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Ans: c

Solution : Mr.Gupta : Mrs. Gupta

Before 7x : 8x
After 5y : 6y

before 7x 8x = 120 15x = 120
x = 8
Mr.Gupta = 7 × 8 = 56
Mrs.Gupta = 8 × 8 = 64
After lossing 6 kg by Mr.Gupta
The ratio becomes 5 : 6

Let Mrs.Gupta loss z kg

300 = 320 – 5z

5z = 20
z = 4 kg
What is the mean proportional between (15 sq 200 ) and (27 – sq 648 ) ?
a) b) 2 c) 3 d) 4
Ans: c
Solution : Mean proportional between a and b = sq ab
= sq (15 200) (27 – sq 648)
= sq (15 x 27) – 15 x sq 648 27 sq 200 – [ sq200 x sq 648)
= sq 405 – 15 x 18 sq 2 27 x 10 sq 2 – [ 10 sq 2 x 18 sq 2]
= sq 405 – 270 sq 2 270 sq 2 – 180 x 2

= sq 405- 360

= sq45

= 3sq 5
To be continued

M. Venkat

Director
MVK Publications
Dilsukhnagar
7671002120