This is part of our
GRE Math Essentials project is the best complement to our
Official GRE Quant Book. It provides a cutting-edge, in-depth overview of all the math concepts to achieve 170 in the Quantitative Reasoning portion of the GRE. The book still remains our hallmark. However, the following chapters will give you a lot of tips, tricks, and shortcuts to make your quant preparation more robust and solid.
1. Find two consecutive odd numbers the difference of whose squares is 296.
Sol. Let the numbers be 2X + 1 and 2X + 3 Then (2X + 3)² – (2X + 1)² = 296 ⇒ X = 36
Hence 2X+ 1 = 2 x 36 + 1 = 73 and 2X + 3 = 75 ⇒ The required numbers are 73 and 75. [Verification. (75)² – (73)²= 5625 – 5329 = 296]
2. A is 29 years older than B, B is 3 years older than C and D is 2 years younger than C. Two years hence A’s age will be twice the united ages of B, C and D. Find their present ages.
Sol. Let D’s age be = X years Then C’s age = (X + 2) years, B’s age = (X + 5) years and A’s age = (X + 34) years. Two yrs hence A’s, B’s, C’s and D’s ages will be X + 36, X + 7, X + 4 and X + 2 years respectively. ⇒ 2 (X + 2 + X + 4 + X + 7) = X + 36 ⇒ X = 2 ⇒ A’s age = 36 yrs; B’s age = 7 yrs, C’s age = 4 yrs; D’s age = 2 yrs.
3. A number consists of three consecutive digits, that in the unit’s place being the greatest of them three. The number formed by reversing the digits exceeds the original number by 22 times the sum of the digit. Find the number.
Sol. Let the hundred’s digit be X. Then the ten’s digit = X + 1 and the unit’s digit = X + 2 ⇒ The number = 100 x X + 10(X + 1) + X + 2 = 111X + 12. The number formed by reversing the digits = 100(X + 2) + 10(X + 1) + X = 111X + 210 ⇒ 111X + 210 – 111X – 12 = 22 (X + 2 + X + 1 + X) ⇒ X = 2. Hence the required number = 234.
4. The crew of a boat can row at the rate of 5 miles an hour in still water. If to row 24 miles, they take 4 times as long as to row the same distance down the river, find the speed at which the river flows.
Sol. Let X miles per hour be the speed of the river. Hence, on equating the times, we get: \(\frac{24}{(5 – X)} = 4 \times \frac{24}{(X + 5)} ⇒ X = 3\) Thus, the river flows at the rate of 3 miles an hour.
5. The area of a rectangle remains the same if the length is increased by 7 meters and the breadth is decreased by 3 meters. The area remains unaffected if the length is decreased by 7 m and breadth is increased by 5 m. Find the dimensions of the rectangle.
Sol. Let the length of the rectangle be X m and breadth of the rectangle = Y m. Area = XY sq. m.
I Case : Length = (X + 7) m and breadth = (Y – 3) m ⇒ Area = (X + 7) (Y – 3) sq. m. ⇒ (X + 7) (Y – 3) = XY or – 3X + 7Y – 21 = 0 ....
II Case : Length = (X – 7) m and breadth = (Y + 5) m ⇒ (X – 7) (Y + 5) = XY or 5X – 7Y – 35 = 0 .... (2) ⇒ Y = 15 and X = 28 Hence length = 28 m and breadth = 15 m Answer.
Note : Assume L = 28 and B = 15 as an option and try checking the conditions given in the problem. You will see that working backward is exceptionally fast in such cases.
6. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them saves Rs. 200 per month, find their monthly incomes.
Sol. Let the monthly income of first person be Rs 9X and the monthly income of second person be Rs 7X. Let the expenditure of first person be 4Y and the expenditure of second person be 3Y. ⇒ Saving of the first person = Rs (9X – 4Y) and solving of second person = Rs (7X – 3Y). Using the given informations, we have : 9X – 4Y = 200 .... (1) and 7X – 3Y = 200 .... X = 200. Hence, the monthly income of first person = Rs. 9 x 200 = Rs. 1800 and the monthly income of second person = Rs. 7 x 200 = Rs. 1400
7. Find two consecutive even numbers such that 1/6th of the greater exceeds 1/10th of the smaller by 29.
Sol. Let the numbers be 2X and 2X + 2 Then (2X + 2)/6 – 2X/10 = 29. ⇒ X = 21 Hence 2X = 430 and 2X + 2 = 432. ⇒ The required numbers are 430 and 432. [Verification. \(\frac{432}{6} – \frac{430}{10} = 72 – 43 = 29\)]
8. A number consists of two digits whose sum is 12. The ten’s digit is three times the unit’s digit. What is the number?
Sol. Let the unit’s digit be X, Then the ten’s digit is 12 – X. ⇒ 3X = 12 – X ⇒ X = 3 Hence the number is 93. [Verification. 9 = 3 x 3; and 9 = 3 = 12]
9. A train travelled a certain distance at a uniform rate. Had the speed been 6 miles an hour more, the journey would have occupied 4 hours less; and had the speed been 6 miles an hour less, the journey would have occupied 6 hours more. Find the distance.
Sol. Let us suppose that X miles per hour is the speed of the train and Y hours is the time taken for the journey. ⇒ Distance travelled = XY = (X + 6) (Y – 4) = (X – 6) (Y + 6) This gives two simultaneous equations. Solving, we get : X = 30, Y = 24 ⇒ Distance = XY = 720 miles
10. A sum of money was divided equally among a certain number of persons; had there been six more, each would have received a rupee less, and had there been four fewer, each would have received a rupee more than he did; find the sum of money and the number of men.
Sol. Let X be the number of persons and Rs Y be the share of each. Then by the conditions of the problem, we have (X + 6) (Y – 1) = XY .......(1) (X – 4) (Y + 1) = XY .......(2). Thus the number of person is X = 24 and the share of each is Y = Rs 5. The sum of money = 5 x Rs 24 = Rs 120.
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