Cetking launches advance shortcut workshops…

Algebra Equations and Polynomials …

Topic this week:
Algebra Equations and Polynomials shortcut workshops

These questions are going to be covered please go through them.

Handout is shared because i want you to see the questions before you come so that we can focus on shortcuts only.

Homework before workshop:
Do basics of simple and quadratic equations or watch basic lectures of Equations on cetking.com

Venue:
All India – Online
Mumbai in class-  Cetking centres in Thane Dadar Vashi

Sample shortcut video

Getting started with Equations

Agenda and topics covered:
Topic covered in workshop.
Simple Equations
Polynomials
Functions & Graphs
Modulus & Inequalities
Maxima & Minima
Special Equations

Introduction

Equations is not just about questions on ages, numbers etc.. there is much more to it… Equations come as a subset of Algebra… We still call the topic equations rather than Algebra or Polynomials so that students study and dont get scared off.

Scope of Algebra Equations and Polynomials shortcuts in CAT & CET and Other exams:

Importance

Importance of Equations in CAT – 5 questions
Importance of Equations in CMAT – 4 ques
Importance of Equations in NMAT – 6 ques
Importance of Equations in SNAP – 6 ques
Importance of Equations in CET – 2 ques

Other workshops coming soon….

Cetking Courses….

CONCEPT TEST (Easy ques): EQUATIONS…

CONCEPT TEST (Easy ques): EQUATIONS as homework before the workshop.

Q.1:
Solve: 2 (3u – v) = 5uv, 2 (u + 3v) = 5uv
(a) u = 2, v = 1
(b) u = 3, v = 2
(c) u = 4, v = 3
(d) none of these

Q.2:
A florist was asked to make a bouquet worth exactly Rs. 1000 with 100 sticks of roses of three colours – Pink, Yellow and Red. While Pink roses cost Rs. 0.50 a stick, Red roses cost Rs. 10.00 per stick and Yellow roses cost Rs. 50.00 per stick. How many Red roses did the florist use in the bouquet?

(a) 1
(b) 5
(c) 80
(d) several combinations are possible

Q.3:
A man has some hens and cow. If the number of heads be 48 and number of feet equals 140, the number of hens will be
(a) 26 (b) 24 (c) 23 (d) 22

Q.4:
A woman sells to the first customer half her stock and half an apple; to the second customer she sells half her remaining stock and half an apple, and so on to the third, and to a fourth customer. She finds that she has now 15 apples left. How many apples did she have before she started selling?

(a) 63 (b) 127 (c) 240 (d) None of these (289)

Q.5:
Find the value of k for which the system of equations
2x + ky = 1, 3x – y = 7 has a unique solution.
(a) k = -2/3 (b) k ≠2/3 (c) k ≠ -2/3 (d) k = 2/3

Q.6:
Ten years ago, Mohan was thrice as old as Ram was but 10 years hence, he will be only twice as old. Find Mohan’s present age.
(a) 60 years
(b) 80 years
(c) 70 years
(d) 76 years

Q.7:
In an examination, a student attempted 15 questions correctly and secured 40 marks. If there were two types of questions (2 marks and 4 marks questions), how many questions of 2 marks did he attempt correctly?
(a) 5 (b) 10
(c) 20 (d) 40
Q.8:
The ages of A, B and C together total 185 years. B is twice as old as A and C is 17 years older than A. Then, the respective ages of A, B and C are
(a) 40, 86, and 59 years
(b) 42, 84 and 59 years
(c) 40, 80 and 65 years
(d) None of these

Q.9:
If the cost of 3 audio and 2 second hand video cassettes is Rs. 350 and that of 2 audio and 3 second hand video cassettes is Rs. 425, then what is the price of a second hand video cassette?

(a) Rs. 110 (b) Rs. 115
(c) Rs. 125 (d) Rs. 140

Q-10:
A two digit number becomes five – sixth of itself when its digits are reversed. The two digits differ by one. The number is:

a) 48 b) 53 c) 54 d) 59 e) 43

Q.11:
The sum of the numerator and denominator of a fraction is 11. If 1 is added to the numerator and 2 is subtracted from denominator, it becomes 2/3. Find the fraction.

a) 4/9 b) 5/3 c) 3/8 d) 5/9 e) 4/7

Q.12:

Sushil planned to drive a distance of x km. after driving 120 km, Sushil stopped for petrol. What fractional part of the trip had Sushil covered when he stopped?

a) x/120 (b) x/((x+120)) (c) 1/((x+120)) (d) ((x-120))/x

Q.13:
A tells B “If you give me Rs. 400, then I shall have 25% more than what you have”. B tells A, “If you give me Rs. 200, then I shall have three -and-a-half times as much as you have”. How much does B have?

(a) Rs. 1200 (b) Rs. 900 (c) Rs. 800 (d) Rs. 1600

Q.14:
A lady went to the market with a few one-rupee notes and a few 20-paise coins. When she returned, she has as many one-rupee notes as she originally and has 20-paise coins and vice versa. She actually came back with about one-third of what she has gone with. How much did she spend in the market?

(a) Rs. 14.40
(b) Rs. 14.70
(c) Rs. 15.50
(d) Rs. 17.40

Q.15:
A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

(1) 2 ≤ x ≤ 6 (2) 5 ≤ x ≤ 8
(3) 9 ≤ x ≤ 12 (4) 11≤ x ≤ 14

Q.16:

In a zoo, there are rabbits and pigeons. If their heads are counted, there are 90 white their legs are 224. Find the number of pigeons in the zoo.

(a) 70
(b) 68
(c) 72
(d) 22

Q.17:
A student was asked to divide a number it by 17/8. Instead, he actually multiplied it by 17/8 and hence got 225 more than the expected answer. What was the expected answer?
(a) 126
(b) 136
(c) 64
(d) None of this

Q.18:
A man’s age is 125% of what it was 10 years ago, but 88 1/3% of what it will be after ten years. What is his present age?
(a) 45 years
(b) 50 years
(c) 55 years
(d) 60 years

Q.19:
The difference between a two – digit number and its reverse (interchanging the digits) is 36. If the difference between the digits is 4 then what is the original number?
a) 18 b) 53 c) 37 d) 29 e) 33

Q.20:
If 2/χ + 3/3y = 9/χy and 4/x + 9/y = 21/χy where χ≠ 0, у ≠ 0, the values of χ and у are, respectively.
3 and 1 (b) 1 and 2 (c) 2 and 3 (d) 1 and 3

1.
2(3u – v) = 5uv – = 5
2(u + 3v) = 5uv + = 5
Putting = x and = y, we get
6x – 2y = 5…..(i)
2x + 6y = 5……(ii)
Multiplying Eq. (i), and Equ. (ii) by 3 and 1 respectively, we have
18x – 6y = 15 and 2x + 6y = 5
Adding these two equations, we get 20 x = 20, i.e. x – 1
From Eq. (i) y = = =
u = = 2 and v = = 1

2.
P + R + Y = 100 ….(i)
0.50P + 10R + 50Y = 1000 ….(ii)
From (i) R = 100 – (P + Y)
0.50 + 10 (100 – P- Y) + 50 Y = 1000
or 0.50 + 100 – 10P – 10 Y + 50Y
= 1000
or 9.5p = 40Y or = =
The florist can use :
80 sticks of Pink = Rs. 40
19 sticks of Yellow = Rs. 950
1 sticks of Red = Rs. 10
100 sticks = Rs. 1000

3.
Let there be hens and y cows
Then, + y = 48 ……(i)
and 2 + 4y = 140 ……(ii)
Solving (i) and (ii), we get = 26.

4.
Suppose she had apples in the beginning.
Sold to the first customer = + =
Sold to the second customer = +
= =
Sold to the third customer = +
= =

Sold to the forth customer = + =
– + + + = 15
– = 15
= 240 + 49 = 289.

5.
For a unique solution to exist we require
≠ k ≠ –

6.
Let Mohan’s present age be x years and Ram’s present age be y years.
Then, according to the first condition,
x-10 = 3(y-10)
or x-3y = -20 (1)
Now, Mohan’s age after 10 years = (x+10) years
Ram’s age after 10 years = (y+10)
:. (x+10) = 2 (y+10)
Or x-2y = 10 (2)
Solving (1) and (2), we get
x = 70 and y = 30
:. Mohan’s age = 70 years and Ram’s age = 30 years.

7.
Suppose number of 2 – mark questions =
Suppose number of 4 – mark questions = y
+ y = 15 => + y = 40 = 10, y = 5.

8.
Let A’s age be x years
B’s age be 2x years
C’s age = (x+17) years
According to the question,
x+2x+(x+17) = 185
:. 4x = 185-17 = 168 :. x=42
:. A’s age = 42 years
B’s age = 84 years
C’s age = 42+17 = 59 years.

9.
3A +2V = 350
2A +3V = 425 A + V = 155 and
A – V = -75
A = 40, V = 115.

10.
A two digit number becomes five – sixth of itself when its digits are reversed.
(10a+b) = 5/6(10b+a).. ..i
The two digits differ by one a – b = 1 …ii
Solving equations answer is 54

11.
The sum of the numerator and denominator of a fraction is 11. If 1 is added to the numerator and 2 is subtracted from denominator, it becomes 2/3. Find the fraction.Using answer choices 3/8.

12.
of the planned distance.

13.
Let A has Rs. and B had Rs. Y in the beginning.
If B gives, Rs. 400 to A, then
(y – 400) 4 – 5y = – 3600 ….(i)

If A gives Rs. 200 to B, then
( – 200) 7 – 2y = 1800 ….(ii)
Solving (i) and (ii), we get
= 600, y = 1200

14.
Originally, the lady had, say, One – rupee notes and y 20 – paise coins.
On returning, she had y one-rupee notes and one – rupee notes and 20-paise conis.
The balance was Rs.
– = ….(i)
=
If we take = 13 and y = 7, then originally she had Rs. 14.40 with her.
On returning, she had Rs. 9.60
The lady spent Rs. 4.80 or a multiple of Rs. 4.80, i.e. Rs. 14.40, which is alternative (a).

15.
Amount of rice bought by the first customer = (X/2 + ½) kgs Amount of rice remaining X – (X/2 + ½) = (X-1)/2 kgs
Amount of rice bought by the second customer 1/2 x (X-2)/2 + ½ = (X+1)/4 kgs
Amount of rice remaining = (X-1)/2 kgs – (X+1)/4 kgs = (X-3)/4 kgs
Amount of rice bought by the third customer ½ x (X-3)/4 +1/2 = (X+1)/8 kgs
As per the information given in the question (X+1)/8 kgs = (X-3)/4 kgs because there is no rice left after the third customer has bought the rice. Therefore, the value of ‘x’ = 7 kgs. Hence, option (2) is the correct choice

16.
R+ P = 90
4R + 2P = 224
R= 22, P = 28.

17.
– 225 –
225 = 136 X 225
= 136

18.
Let the present age be x years.
Then, 125% of (x-10) = x
and 83 1/3% of (x+10) = x
:. 125% of (x-10) = 83 1/3% of (x+10)
or 5/4 (x-10) = 5/6 (x+10)
or 5/4x-5/6x = 50/6+50/4
or 5x/12 = 250/12 or x = 50 years.

19.
The difference between a two – digit number and its reverse (interchanging the digits) is 36.
(10a+b) – (10b-a) = 36
9(a-b) = 36 => (a-b)= 4 … (i)
If the difference between the digits is 4
(a-b) = 4 … (ii)
Hence no solution from equations. Solve the question using answer choices.

20.
Multiplying each equation throughout by , we get 3 + 2y = 9 and 9 + 4y = 21
On solving these equations, we get = 1, y = 3.

Equations in CAT & CMAT Exams