Posts

Showing posts with the label ssc cgl

The ratio of Boys and Girls in a group is 7 : 6. If 4 more boys join the group and 3 girls leave the group, then the ratio of boys to girls becomes 4 : 3. What is the total number of boys and girls initially in the group?

A. 78 B. 104 C. 117 D. 91 SSC CGL 2019 Tier - 1 4 March 2020 Shift 2 Solution: Let the number of boys and number of girls initially be x and y respectively. It is given that, x/y = 7/6 6x - 7y = 0 ............(1) If 4 more boys join the group and 3 girls leave the group Number of boys in the group = x + 4 Number of girls in the group = y - 3 New ratio is (x + 4)/(y - 3) = 4/3 3x + 12 = 4y - 12 3x - 4y = -24 ............(2) Multiply equation 2 by 2 and subtract from equation 1, we get y = 48 Put y = 48 in eq 1, we get, 6x - 7 * 48 = 0 x = 56 The number of boys and girls initially in the group = x + y = 48 + 56 = 104 Hence, (B) is the correct answer. This question is based on the topic of Ratio and Proportion .
Post a Comment
Read more

A trader bought two articles for Rs. 490. He sold one at a loss of 20% and the other at a profit of 16%. If the selling price of both the articles is same, then the cost price (in Rs.) of the article sold at 20% loss will be:

A. 300 B. 280 C. 310 D. 290 SSC CGL 2022 13 August 2021 Shift 1 Solution: Let the articles be A1 and A2 sold at 20% loss and 16% profit respectively. Let the Cost Price (CP) of the A1 for the trader be = x Then the Cost Price (CP) of the A2 for the trader will be = 490 - x It is given that the Selling Price (SP) of A1 and A2 are the same. Let the Selling Price be = y It is given that, On sale of A1, Loss % = 20% On sale of A2, Profit % = 16% We know that, Loss % = ((CP - SP)/CP) * 100 .......(1) Profit % = ((SP - CP)/CP) * 100 ........(2) Putting values in eq. 1 for A1 and in eq. 2 for A2 20 = ((x - y)/x) * 100 .......(3) 16 = ((y - (490 - x))/(490 - x)) * 100 ....(4) From eq. 3, we get 20/100 = 1 - y/x y/x = 1 - 20/100 y/x = 80/100 y = 4x/5 ....(6) Substitute eq. (5) in eq. (4) 16 = ((4x/5 - (490 - x))/(490 - x)) * 100 16/100 = (4x/5)/(490 - x) - 1 4x/(5(490 - x)) = 16/100 + 1 4x/(5(490 - x)) = 116/100 4x/(5(490 - x)) = 29/25 100x = 29 * 5 * (490 - x) 100x = 71050 - 145x 245x...
Post a Comment
Read more

If the number 1005x4 is completely divided by 8. Then the smallest integer in place of x will be:

A. 0 B. 1 C. 4 D. 2   SSC CGL 2019 Tier 1 3 March 2022 Shift 1 Maths Paper Question Solution: Divisibility Rule of 8: If a number is divisible by 8, then its last three digits should be divisible by 8. In the given number 1005x4, the last four digits are 5x4 Let us check by using the options: using x = 0 divide 504 by 8, and we will get 63. Here, putting x = 0 makes the given digit divisible by 8. Therefore, (A) is the correct answer.
Post a Comment
Read more