Solved Examples - Arithmetic - GMAT PDF Download

(Section - 1)
Ques 1: 39 -(25 -17 )
Ans: 39 - (25 - 17) =
39 - 8 = 31
Tip: You could distribute the minus sign
(39 — 25 + 17) if you prefer, but our method is less prone to error.

Ques 2: 3 (4 - 2 )/ 2
Ans: 3 x (4 - 2) / 2 =
3 x (2) / 2 =
6 / 2 = 3

Ques 3: 15 x 3/9
Ans: 15 x 3 / 9 =
45 / 9 = 5

Ques 4: (9 - 5) - (4 - 2)
Ans: (9 - 5) - (4 - 2) =
(4) - (2) = 2

Ques 5: 14 -3 (4 -6 )
Ans: 14 - 3 (4 - 6 ) =
14 - 3(-2) =
14 + 6 = 20

Ques 6: -5 x 1 / 5
Ans: -5 x 1 + 5 =
-5 + 5 = - l

Ques 7: (4)(—3)(2)(— 1)
Ans: ( 4 ) (-3 )( 2 )(- l) = 24
Tip: To determine whether a product will be positive or negative, count the number of negative terms being multiplied. An even number of negative terms will give you a positive product; an odd number of negative terms will give you a negative product.

Ques 8: 5 - (4 - (3 - (2 - 1)))
Ans: 5 - (4 - (3 - (2 - 1))) =
5 - (4 - (3 - 1)) =
5 - ( 4 - 2 ) =
5 - (2) = 3
Tip: Start with the inner-most parentheses and be careful about the signs!

Ques 9: -4 (5 )-1 2 /(2 + 4)
Ans: - 4 ( 5 ) - 1 2 / ( 2 + 4) =
-20 - 12/(6) =
-20 - 2 = -22

Ques 10: 17(6)+ 3(6)
Ans: 17 (6 ) + 3 (6 ) =
102 + 18 = 120
Alternatively, you could factor the 6 out of both terms.
17 (6 ) + 3 (6 ) =
6 (17 + 3) =
6(20)= 120

(Section - 2)
Ques 11: —12 x 2/(—3) + 5 =
Ans: -12 x 2/ (-3) + 5 =
- 24 / ( - 3 ) + 5 =
8 + 5 = 13

Ques 12: 32/(4 + 6 x 2) =
Ans: 32 /(4 + 6 x 2) =
32 /(4 + 12) =
32 /(1 6 ) = 2

Ques 13: —10 — (—3)² =
Ans: - 10 - (-3)²  =
- 10 - (9) = - 1 9

Ques 14: -5² =
Ans:- 5² =
- (5²) = - 25

Ques 15: - 2³/2 =
Ans: - 2³/2 =
- 8/2 = - 4

Ques 16: 5³- 5² =
Ans: 5³- 5² =
125 - 25 = 100

Ques 17: 5⁽²⁺¹⁾ + 25 =
Ans: 5⁽²⁺¹⁾  + 25 =
5³ + 25 = 125 + 25 = 150

Ques 18: (-2)³- 5² + (-4)³ =
Ans: (-2)³ - 5² + (-4)³ =
( - 8) - 25 + (-64) =
-33 - 6 4 = -9 7

Ques 19: 5(1)+ 5(2)+ 5(3)+ 5(4) =
Ans: 5(1) + 5(2) + 5(3) + 5(4) =
5 + 10 + 15 + 20 = 50
Alternatively, you could factor a 5 out of each term.
5(1) + 5(2) + 5(3) + 5(4) =
5(1 + 2 + 3 + 4) =
5(10) = 50

Ques 20: 3 x 99 - 2 x 99 - 1 x 99
Ans: 3 x 99 - 2 x 99 - 1 x 99 =
297 - 198-99 =
99 - 99 = 0
Alternatively, you could factor 99 out of each term.
3 x 99 - 2 x 99 - 1 x 99 =
99(3-2-1) =
99(0) = 0

Section - 3
Ques 21: πr²- (2πr + πr²)
Ans: πr²- (2πr + πr²)
πr²- 2πr + πr²
1πr² - 2πr- 1πr² 
0 - 2πr = - 2πr

Ques 22: 1 + 2 √4 + 2√2
Ans: 1 + 2(2)+ 2√2
1+4 + 2 √2
5 + 2√2

Ques 23: 12xy  - 6(xy)² + (2xy)²
Ans: 12xy² - 6(xy)² + (2xy)² =
12xy² - 6x²y² + 2²x²y² =
12xy² - 6x²y² + 4x²y² =
12xy² +(-6+4)x²y² =
12xy² -2x²y² =
12xy² -2x²y² 

Ques 24: 3π + xπ-2π
Ans: 3π — 2π + xπ =
(3 — 2)π + xπ =1π + xπ =
(1 + x)π = (x + 1)π

Ques 25:  √2 + x√2 - 2√2
Ans: √2 + x√2 - 2√2
1√2 + x√2 - 2√2
(1 + x - 2) √2
(-1 + x )√2 = (x - 1)√2

Ques 26:  12xy- (6x + 2y)
Ans: 12 xy — (6x + 2y) =
12xy — 6x—2y
Note that we cannot actually combine any of the terms.

Ques 27:  3x - (3x + 5 - (2x - 3))
Ans: 3x - (3x + 5 - (2x - 3))
3x - (3x + 5 - 2x + 3)
3x - (3x - 2x + 5 + 3)
3x - (1x + 8) =
3 x - 1x - 8 =
2x - 8

Ques 28:  π²r² - πr+ 2πr² + π(r²) + (πr)² + 2πr
Ans: π²r² - πr+ 2πr² + π(r²) + (πr)² + 2πr
π²r² - πr+ 2πr² + πr² + π²r² + 2πr
(1π²r² +1π²r²⁾+ (2πr² -1πr) + (2πr² + 1πr²)
2π²r² +1πr+3πr²

Ques 29: 2x² - (2x)² - 2² - x²
Ans: 2x² - (2x)² - 2² - x²
2x² - 2²x² - 2² - 1x²
2x² - 4x² - 1x² - 4
(2 - 4 - 1)x² - 4
(-3)x² - 4
-3x² - 4

Ques 30: 4x² + 2x - (2√x )²
Ans: 4x² + 2x - (2√x )²
4x² + 2x - 2²(√x )²
4x² + 2x - 4x
4x² + (2 - 4)x
4x² + (-2)x
4x² +2x

Section - 4
Ques 31: 3 (5 - y )
Ans: 3 (5 -y ) = 3 x 5 + 3 x (-y) = 15-3 y

Ques 32 : -(a-b)
Ans: The minus sign in front of the left parenthesis should be interpreted as: -1 times the expression {a - b).
Because (-1) x a = - a and (-1) x (-b) = b we have:
-(a - b) = ( - 1) x (a — b) = —a + b

Ques 33:  (2x + y)z
Ans: This problem requires us to distribute from the right. Ordinarily, we think of the Distributive Property in this form:
a(b + c) = ab + ac
It is also true that:
(b + c)a = ba + ca
One way to justify this is to note that (b + c)a = a(b + c) by the Commutative Property of Multiplica-tion, which states that the order in which we multiply numbers does not matter. The GMAT sometimes disguises a possible distribution by presenting it in this alternate form.
In this problem, (2x + y)z = z(2x + y).
z(2x + y) = z x 2x + z x y = 2 xz + yz

Ques 34: √3(√2+√3)
Ans: The first step here is to distribute as normal:
√3(√2+√3)= √3√2+√3√3
However, it is best to then simplify this answer. When multiplying together two square roots we can simply multiply together the two numbers under a single square root sign (more on this later):
√a√b=√ab
Therefore, √3√2 = >√3 x 2 = √6 . The second term √3√3 can be dealt with in the same manner, or we can note that it is 3 by definition of the square root. Adding these two terms together gives us the simplified answer.

35. 5.2r(2t- 10s)
Ans: When distributing more complicated expressions, we should remember to multiply out numbers and combine any copies of the same variable. Here we have:
5.2r(2t- 10s) = (5.2r)(2t) - (5.2r)(10s)
Because 5.2 x 2 = 10.4 and 5.2 x 10 = 52, this simplifies to 10.4rt- 52rs.

36. (-3.7x + 6.3)10²
Ans: In this problem we must both distribute from the right and multiply out the numerical expressions. Multiplying a number by 102 = 100 is the same as moving the decimal point two places to the right.
Therefore
(-3.7x + 6.3)10² = -370x + 630

37. 6k²l(k - 21)
Ans: The first step in this computation is to distribute normally:
6k²l x k - 6k²l x 2l
In the first term, we then multiply together the k² and the k to get 6k³l, because k² x k = k³. For the second term, we need to multiply together the 6 and 2 as well as the two copies of l, so we have  =6k²l x 2l = 12k²l². Putting it all together:
6k²l x k - 6k²l x 2l = 6k³l- 12k²l²

38. -√3(-n√12+√27)
Ans: After distributing, we have:
√3(-n√12+√27)= n√3√12 -√3√27
We should then simplify the terms√3√12 and √3√27 using the fact that we can multiply under the square root sign (or more formally, √a√b= √ab).
The first term is √3 √12 = √36 = 6 . The second term involves the multiplication 3 x 27 = 81. The square root of 81 is 9.
-√3(-n√12+√27 = n√3 √12-√3 √27 = 6n - 9

39. d(d² - 2d +1)
Ans: Even though there are three terms inside the parentheses, distribution works exactly the same. Multiply d by every term in the parentheses.
d(d² - 2d +1) = (d x d²)-(d x 2d) + (d x 1)

40. xy²z(x²z+ yz² - xy²)
Ans: The term xy²z on the outside of the parentheses must be multiplied by each of the three terms inside the parentheses. We should then simplify the expression as much as possible. Taking one term at a time, the first is xy²z x x²z = x³y²z², because there are three factors of x, two factors of y, and two factors of z. Similarly, the second term is xy²z x yz2 = xy³z³ and the third is xy²z x (-xy²) = - x²y⁴z. Adding these three terms together gives us the final answer.