# GRE Practice Test - Numeric Entry GRE Practice Test 3

Q. 1

Find r when C(n,r) = 35 and P(n,r) = 840

• EXPLAINATION: From the given conditions, we get C(n,r) = n!/[r!(n-r)!] = 35...(1) P(n,r) = n!/(n-r)! = 840...(2) Dividing (2) by (1), we get n!/(n-r)!/n!/[r!(n-r)!] = 840/35 r! = 24 r! = 1*2*3*4 = 24 r = 4

Q. 2

If (n+1)! = 30*(n-1)!, then find n.

• EXPLAINATION:(n+1)! = 30*(n-1)! (n+1)*n*(n-1)! = 30(n-1)! (n+1)*n = 30 n^2 + n - 30 = 0 n^2 +6n-5n - 30 =0 n(n+6) - 5(n+6) = 0 n = -6, 5 Since n cannot be negative, n = 5

Q. 3

Find x if (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) are the vertices of an equilateral triangle in the first quadrant.

• EXPLAINATION: Let the vertices (0,0), (3, sqrt(3)) and (x, 2*sqrt(3)) be A, B and C respectively. Since the triangle is equilateral, we have AB = BC = CA AB^2 = BC^2 = CA^2 (3-0)^2+(sqrt(3)-0)^2 = (3-x)^2 + (2*sqrt(3) - sqrt(3))^2 = (x-0)^2 + (2*sqrt(3)-0)^2 9 + 3 = 9 - 2x + x^2 +3 = x^2 + 12 Hence, x^2+12 = 12 x = 0 [AB^2=AB*AB]

Q. 4

C can complete the work in 18 days if he works alone. B takes 3 days lesser than C and A takes 5 days lesser than B to complete the work. On which day will they complete the work if C works for the initial two days only?

• EXPLAINATION: C takes 18 days to complete the work. B takes 3 days lesser. Hence, B takes 18-3 = 15 days. A takes 5 days lesser than B. Hence, A takes 15-5 = 10 days. Work done by A in one day = 1/10 Work done by B in one day = 1/15 Work done by C in one day = 1/18 Work done by the three of them in two days = 2(1/18+1/10+1/15) = 2(5+9+6)/90 = 2*20/90 = 4/9 Remaining work = 1-4/9 = 5/9 Work done by A and B in one day = 1/10+1/15 = (3+2)/30 =1/6 Time taken by A and B to complete the remaining work = 5/9*6/1 = 10/3 = 3.33 The work in completed on 2+3.33 = 6th day

Q. 5

If A and B are two mutually exclusive events and P(A) = 2/3 and P(B) = 4/9, then find the probability of the occurrance of A and B together.

• EXPLAINATION: Since A and B are mutually exclusive events, they do not occur together at all. Hence, the required probability is zero.

Q. 6

Find the diagonal of a cuboid whose volume is 144 cc and the dimensions of its base are 12cm and 4cm.

• EXPLAINATION: The length (l) and breadth (b) of the base are 12 cm and 4 cm respectively. Height (h) of the cuboid = volume/(l*b) = 144/(12*4) = 3 cm Diagonal of the cuboid = sqrt(l^2+b^2+h^2) = sqrt(12^2+4^2+3^2) = sqrt(144+16+9) = sqrt(169) = 13 cm The diagonal of the cuboid is 13 cm. [l^2=l*l]

Q. 7

Find the value of k for which both the equations 12x^2+4kx+3=0 and (k+1)x^2-2(k-1)x+1=0 have equal real roots. [x^2=x*x]

• EXPLAINATION:Since the two equations have equal real roots, their discriminants are 0. For 12x^2+4kx+3=0, D = 0 D^2 = 0 b^2-4ac=0 (4k)^2 - 4*12*3 = 0 16k^2 - 144=0 k^2 = 144/16 = 9 k = -3, 3 For (k+1)x^2-2(k-1)x+1=0 b^2-4ac=0 [2*(k-1)]^2-4*(k+1)*1=0 4(k^2-2k+1)-4(k+1)=0 4[k^2-2k+1-k-1]=0 k^2 - 3k = 0 k(k-3)=0 k=0,3 Hence, we have k = 3 [k^2=k*k]

Q. 8

What percent is the first number of the second if the first number is 120% of the third and the second number is 150% of the third?

• EXPLAINATION:Let the first, second and the third numbers be x, y and z respectively. x = 120% of z = 120/100*z y = 150% of z = 150/100*z x/y*100 = (120z/100)/(150z/100)*100 = 80% x is 80% of y.

Q. 9

The difference between two numbers is 2. Each number is less than 14 and their sum is greater than 22. Find the greater number of the two.

• EXPLAINATION:Let the two numbers be x and y such that x < y. y – x = 2, x < 14, y < 14, x + y > 22 y = x + 2, x < 14, x + 2 < 14, x + x + 2 > 22 x < 14, x < 12, 2x + 2 > 22 x < 12, x +1 > 11 x < 12, x > 10 Hence, x can be 11 and the corresponding value of y is 13. The two numbers are 11 and 13.

Q. 10

The sum of the squares of the two positive numbers is 68 and the square of their difference is 36. Find the sum of the numbers.

• EXPLAINATION: Let the numbers be x and y, x>y x^2+y^2=68 and (x-y)^2=36 (x-y)^2 = x^2+y^2-2xy 36 = 68-2xy 2xy=68-36=32 xy=32/2=16 (x+y)^2=x^2+y^2+2xy = 68+2*16 =68+32=100 x+y = sqrt(100) = 10 [x^2=x*x]

Q. 11

The difference between the time when the lightening was seen and the time when the thunder was heard is 10 seconds. Sound covers a distance 330 meters in one second. Find the distance of the thundercloud from the point of observation in meters.

• Answer: 3300 meters
• EXPLAINATION:Distance = speed*time = 330*10 = 3300 meters

Q. 12

Six points lie on a circle. How many cyclic quadrilaterals can be formed by joining the points?

• EXPLAINATION: Since, the points lie on a circle, all the possible quadrilaterals shall be cyclic. Number of quadrilateals = C(6,4) = 6!/(2!4!) = 6*5/2 = 15 Hence, 15 cyclic quadrilaterals can be drawn.

Q. 13

When a number 'a' is increased by 17, it equals 60 times its reciprocal. How many values of 'a' are possible?

• EXPLAINATION:According to the conditions, we have a+17=60*1/a a^2+17a-60=0 a^2 +20a-3a-60=0 a(a+20)-3(a+20)=0 (a-3)(a+20)=0 a=3, -20 When a = 3, a+17 = 3+17 = 20 = 60*1/3 When a = -20, a+17 = -20+17 = -3 = 60*(-1/20) Hence, the two values of 'a' are valid. There are two possible values of 'a'. [a^2=a*a]

Q. 14

The tax on a commodity decreases by 10% and the consumption increases by 20%. What is the percentage increase in the revenue?