Exam day revision

04. 3 PERSONS ROY , PAUL & SINGH TOGETHER
LI NEAR E QU AT I ON S
HAVE ` 51 . PAUL HAS 4 LESS THAN ROY &
01. SUM OF TWO NUMBERS I S 52 AND THEIR SINGH HAS 5 LESS THA N ROY . THEY HAVE
DIFFERENCE IS 2 . TH E NUMBERS ARE A. 20,16,15 B. 15,20,16 C 25,11,15
A. 17,15 B. 12,10 C. 27,25 METHOD 1
Roy Paul Singh TOTAL
x+y = 52
x x4 x5 3x9 = 51
xy = 2
x = 20
2x = 54
Roy Paul Singh
x = 27
` 20 ` 16 ` 15
y = 25 ans 27,25
USE OPTIONS METHOD 2
Option C satisfied bot h the conditions USE OPTIONS
Option A satisfies all the conditions
02. DIAGONAL OF RECTANGLE IS 5 AND ONE OF

ITS SIDE IS 4 CM . A REA IS 05. MONTHLY INCOME OF TW O PERSONS ARE IN
A. 20 B. 12 C. 10 RATIO 4 :5 AND THEIR MONTHLY EXPENSES

ARE IN RATIO 7 :9 . IF EACH SAVES 50
let other side be x
PER MONTH THEIR MONTHLY INCOMES
By PYTHAGORAS THEOREM ,
A. 500,400 B. 400,500 C. 300,600
x 2 + 16 = 25
x = 3 METHOD 1
AREA = 4x3 = 12 I II
INCOME 4x 5x
03. PRODUCT OF TWO NUMBE RS IS 3200 AND EXPENSES 7y 9y
THE QUOTIENT WHEN THE LARGER NUMBE R SAVINGS 4x7y 5x9y = 50
IS DIVIDED BY THE SM ALLER IS 2 . THE 5x9y = 50
NUMBERS ARE 4x7y = 50
A. 16,200 B. 160,20 C. 60,30 D. 80,40 x2y = 0 x = 2y
subs in (1)
METHOD 1
y = 50 , x = 100
x.y = 3200 …. (1)
INCOMES 400 , 500
x/y = 2
x = 2y …. (2) METHOD 2
subs in (1) USING OPTIONS
2y 2 = 3200 OPTION b
y = 40 INCOME 400 500
x = 80 ans 80,40 RATIO 4 5 SAT ISFIE D
METHOD 2 SAVINGS 50 50
USE OPTIONS EXPENSES 350 450
Option d satisfied both the conditions RATIO 7 9 SAT ISFIE D

,06. THE WAGES OF 8 MEN A ND 6 BOYS IS ` 33. 08. FOR A CERTAIN COMMOD ITY , DEMAND

IF 4 MEN EARN ` 4.50 MORE THAN 5 BOY S EQUATION GIVING DEMA ND D IN KG FOR A

DETERMINE THE WAGES OF EACH MAN AND PRICE P IN RUPEES / KG IS
BOY D = 100(10P).
A. ` 1.50,3 B. ` 3,1.50 C. ` 2.50,2 THE SUPPLY EQUATION GIVING THE SUPPLY
D. ` 2,2.50 S IN KG FOR A PRICE P IN RUPEES/ KG IS
METHOD 1 S = 75(P3) .
wages of man = x/- , boy = y/- THE MARKET PRICE IS SUCH THAT DEMAND
8x+6y = 33 ….. (1) EQUALS TO SUPPLY . F IND THE MARKET
4x5y = 4.50 x 2 PRICE AND QUANTITY
8x10y = 9 ….. (2) A. 7/-,300KG B. 7/-,350KG
SOLVING (1) & (2) C. 7/KG,200KG
16y = 24
METHOD 1
y = 1.50 ,x = 3 OPTION A.
d = 100(10p)
METHOD 2 s = 75(p3)
USING OPTIONS , d = S
OPTION b satisifies both the conditions 1000 – 100p = 75p  225
175p = 1225
07. THE DEMAND AND SUPPL Y EQUATIONS OF A p = 7/-
CERTAIN COMMODITY AR E 4Q + 7P = 17 AND d = 100(107) = 300
P = Q + 7 s = 75(73) = 300
3 4 , P = PRICE , Q = DEMA ND
ans 7/- , 300kg
EQUILIBRIUM PRICE AN D QUANTITY
A. 2, B. 3, C. 5,
METHOD 2

METHOD 1 At p = 7/-
d = 100(107) = 300 kg
Equilibrium price is the point where
s = 75(73) = 300 kg
demand and supply functions meet i.e. its
d = s I S SATI SF IE D
the point of intersection of the above two
q = 300 kg
lines and hence solve .
SO OPTION a.
4q + 7p = 17
4q  12p =21
19p = 38
p = 2
sub in (1)
q = ¾ ans 2,

METHOD 2
USING OPTIONS ,
OPTION a satisifies both the equations .
THIS IS FASTER

, LI NEAR E QU AT I ON S




01. 10 YEARS AGO , FATHE R WAS 4 TIMES O F

HIS SON AGE . TEN YE ARS HENCE , THE 02. 5 YEARS AGO , FATHER WAS 7 TIMES OF HIS

FATHER WAS TWICE THE SON AGE . THEIR SON AGE . 5 YEARS FR OM NOW , THE
PRESENT AGES ARE FATHER WAS 3 TIMES THE SON AGE .
A. 50,20 B. 60,20 C. 55,25 FATHER’S AGE IS

A. 40 B. 30 C. 35
METHOD 1
FATHER SON
FATHER SON
5 YEARS AGO 7x x
10 YEARS AGO 4x x
PRESENT 7x+5 x+5
PRESENT 4x+10 x+10
5 YRS HENCE 7x+10 x+10
10 YRS HENCE 4x+20 x+20
7x+10 = 3(x+10)
F A THER IS T WICE THE AGE OF SON
x = 5
4x+20 = 2(x+20)
PRESENT AGE 40
x = 10
PRESENT AGES 50 20 METHOD 2
Using Options in this case is not
METHOD 2
recommended as both the ages are not
USING OPTIONS ,
given in the option .
Option a. satisfies both the conditions
Let me show you
FATHER SON
Using OPTION a.
PRESENT 50 20
FATHER SON
10 YEARS AGO 40 10
5 YEARS BACK 35 5
F a ther is 4 times s ons a ge - SA TI SF IE D
PRESENT 40 10
10 YRS HENCE 60 30
5 YEARS HENCE 45 15
F a ther is tw ic e the s on’ s age – SA TI SFIE D
FATHER IS 3 TIMES THE AGE OF SON –
CON DI TI ON I S SATI SF IE D



03. THE AGE OF A PERSONS IS TWICE THE SUM

OF THE AGES OF HIS T WO SON’S AND FIVE

YEARS AGO HIS AGE WA S THRICE THE SUM

OF THEIR AGES . FIND HIS PRESENT AGE

A. 60 B. 52 C. 51 D. 50
FATHER SON1+SON2
PRESENT 2x x
5 YRS AGO 2x5 x(5+5)
x10
F ATHER IS T HR ICE THE SUM OF SON ’ S AGE S

2x5 = 3(x10)
x = 25
Father’s present age = 50 YRS

, LI NEAR E QU AT I ON S
04. Y IS OLDER BY X BY 7 YEARS. 15 YEARS

BACK , X AGE WAS ¾ O F Y’S AGE . THEIR 01. A NUMBER CONSISTS OF TWO DIGITS . THE

PRESENT AGES ARE DIGIT IN THE TEN’S P LACE IS 3 TIMES THE

A. 36,43 B. 50,43 C. 43,50 D. 40,47 DIGIT IN THE UNIT PLACE . IF 54 IS
METHOD 1 SUBTRACTED FROM THE NUMBER , THE
X Y DIGITS GET REVERSED . THE NUMBER IS
PRESENT AGE x x+7 A. 39 B. 92 C. 93 D. 94
15YRS BACK x15 x8
x15 = 3 Option 39 92 93 94
x8 4 Cond 1    
4x60 = 3x24 Cond 2  54
x = 36 , 39 
x+7 = 43 , option a option c
METHOD 2
USING OPTIONS 02. A NUMBER CONSISTS OF TWO DIGITS . THE

Y IS OLDER BY X BY 7 YEARS DIGIT IN THE TEN’S P LACE IS TWICE THE
36,43 50,43 43,50 40,47 DIGIT IN THE UNIT PL ACE . IF 18 IS
    SUBTRACTED FROM THE NUMBER , THE

DIGITS GET REVERSED . THE NUMBER IS
15 YEARS BACK a. 63 b. 42 c. 84 d. 48
21, 28 28,35 25,32
X AGE WAS ¾ OF Y’S A GE Option 63 42 84 48
   Cond 1    
Cond 2 18  18
option a 45  24 
option c
05. IF TWICE THE SON’S A GE IS ADDED TO THE

FATHER’S AGE , SUM I S 70 . BUT IF TWICE 03. A NUMBER BETWEEN 10 AND 100 IS 5 TIMES

THE FATHER’S AGE IS ADDED TO SON’S AGE , THE SUM OF DIGITS . IF 9 IS ADDED TO IT

THE SUM IS 95 . THEIR AGES ARE THE DIGITS GET REVER SED . THE NUMBER
a. 40,15 b. 15,40 c. 25,30 IS
METHOD 1 A. 54 B. 53 C. 45 D. 55
FATHER AGE = x YRS
SON AGE = y YRS Option 54 53 45 55
x+2y = 70 Sum of

2x+y = 95 Dig its 9 8 9 10
(1) + (2) 3x+3y = 165 Cond 1 x 5 x 5 x 5
x+y = 55 =45  =40 =45 
(2)(1) xy = 25
2x = 80 Cond 2 45
x = 40 , y = 15 + 9
54 
AGES ARE : 40,15
option c