Karnataka SSLC Mid Term Question Paper 2025 Maths

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Papers
Mid Term

PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ
6 ®æà AvÜxÃÜÓæ¤, ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá & 560 003

KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
6th Cross, Malleshwaram, Bengaluru – 560 003

2025-26 ®æà ÓÝ騆 GÓ….GÓ….GÇ….Ô. A«ÜìÊÝÑìPÜ ±ÜÄàûæ
S.S.L.C. MID-TERM EXAMINATION
FOR THE YEAR – 2025-26
…Œ⁄æ⁄fl : V⁄{}⁄
Subject : MATHEMATICS
( BMW«ŒÈ »⁄·¤®⁄¥¿»⁄fl / English Medium )

ËÐÜ¿á ÓÜíPæàñÜ : 81-E Subject Code : 81-E

©®ÝíPÜ : 16. 09. 2025 ] [ Date : 16. 09. 2025
ÓÜÊÜá¿á : ¸æÙÜWæY 10-30 Äí¨Ü ÊÜá«ÝÂÖÜ° 1-45 ÃÜÊÜÃæWæ ] [ Time : 10-30 A.M. to 1-45 P.M.
WÜÄÐÜu AíPÜWÜÙÜá : 80 ] [ Max. Marks : 80

General Instructions to the Candidate :
1. This Question Paper consists of 38 questions.
2. Follow the instructions given against the questions.
3. Figures in the right hand margin indicate maximum marks for the questions.
4. The maximum time to answer the paper is given at the top of the question paper.
It includes 15 minutes for reading the question paper.

I. Four alternatives are given for each of the following questions / incomplete

statements. Choose the correct alternative and write the complete answer

along with its letter of alphabet. 8×1=8

1. The degree of the polynomial p ( x ) = 2 x 3 + 5x – 6 is

(A) 2 (B) 3

(C) 5 (D) 6

MID/T-25-26/28 [ Turn over

81-E 2

2. If the pair of linear equations a1x + b1y + c1 = 0 and a 2 x + b 2y + c 2 = 0

have unique solution, then the correct relation among the following is
a1 b1 a1 b1 c1
(A) ≠ (B) = =
a2 b2 a2 b2 c2

a1 b1 a1 b1 c1
(C) = (D) = ≠
a2 b2 a2 b2 c2

3. The quadratic equation among the following is

(A) x 4 − 3x + 2 = 0 (B) 2x + 3 = 0

(C) x 2 − 5x + 6 = 0 (D) 2x 3 + 7 x + 1 = 0

4. The common difference of the arithmetic progression 100, 93, 86 ... is

(A) – 7 (B) 7

(C) 3 (D) 5

5. If the n th term of an arithmetic progression is 5n + 3, then the 3 rd term of

the progression is

(A) 11 (B) 12

(C) 13 (D) 18
AX
6. In the given figure, if XY || BC, then is equal to
AB

AX AX
(A) (B)
AY XB
AY AC
(C) (D)
AC AY

MID/T-25-26/28

3 81-E

7. The number of zeroes of the polynomial p ( x ) in the given graph is

(A) 0 (B) 1

(C) 2 (D) 3

8. If one root of the quadratic equation 2 x 2 – ( k + 1 ) x + 3 = 0 is 1, then the

value of k is

(A) 0 (B) 1

(C) 3 (D) 4

II. Answer the following questions : 8×1=8

9. State ‘Fundamental Theorem of Arithmetic’.

10. How many solutions does the pair of linear equations in two variables have

if they are inconsistent ?

11. Write the formula to find the sum of first n positive integers.

12. Write the two conditions for similarity of two polygons having same number

of sides.

13. If the product of the zeroes of the polynomial p ( x ) = 2 x 2 – 3x + k is 3,

then find the value of k.

MID/T-25-26/28 [ Turn over

81-E 4

14. If the sum of three consecutive terms of an arithmetic progression is 21,

then find the second term.

15. 3 bats and 2 balls together cost Rs. 700. If the cost of a ball is Rs. 50, then

find the cost of a bat.
AB 2
16. Given ∆ ABC ~ ∆ DEF. If = and BC = 4 cm, then find the measure
DE 3

of EF.

III. Answer the following questions : 8 × 2 = 16

17. What is a composite number ? Which is the composite number among 23

and 24 ?

18. Prove that 2 + 5 is an irrational number.

19. Solve the given pair of linear equations by elimination method :

2x + y = 8

3x – y = 7

OR
a1 b1 c1
On comparing the ratios , and , find whether the lines
a2 b2 c2

representing the pair of linear equations

9x + 3y + 12 = 0 and 18x + 6y + 24 = 0, intersect at a point or parallel or

coincident.

20. Find whether 77 is a term of the arithmetic progression 1, 5, 9, 13, .......

using formula.

OR

In an arithmetic progression, the common difference is 3 and its tenth term

is 32. Find the first term of the progression.

MID/T-25-26/28

5 81-E

21. In the given figure, DE || OQ and DF || OR. Show that EF || QR.

22. In ∆ ABC, DE || BC. If AD = x, DB = x – 4, AE = x + 4 and EC = x – 2, then

find the value of x.

23. A teacher has 96 pens and 128 pencils. Find the maximum number of

students to whom pens and pencils can be equally distributed so that no

pens and pencils are left out.

24. “Two students A and B together have 25 marbles. Each of them lost

5 marbles. The product of the marbles they have now is 50.” Form a

quadratic equation in the standard form to find the number of marbles they

had.

IV. Answer the following questions : 9 × 3 = 27

25. Prove that 3 is an irrational number.

OR

MID/T-25-26/28 [ Turn over

81-E 6

Find the HCF and LCM of the numbers 510 and 92 by prime factorisation

method and verify that LCM × HCF of those numbers = product of the

numbers.

26. If α and β are the two zeroes of the polynomial p ( x ) = x 2 + 3x + 1, then

find the values of
1 1
i) +
α β

ii) α 2 + β2 .

OR

One of the zeroes of the polynomial p ( x ) = 2 x 2 – 6x + k is twice the other.

Find the value of k.

27. A train travels 480 km at a uniform speed. If the speed had been 10 km/h

more, it would have taken 4 hours less for the same journey. Find the speed

of the train.

OR

The altitude of a right angled triangle is 7 cm less than its base. If the

hypotenuse is 13 cm, then find the other two sides.

28. Find the discriminant of the quadratic equation x 2 + 7x + 12 = 0. If real

roots exist for the equation, then find them.

29. Five years hence, the age of A will be three times that of his son. Five years

ago, A’s age was seven times that of his son. Find their present ages.

30. Find the sum of all the integers between 1 and 100, all of which leave a

remainder 1 when divided by 5 using formula.

OR

Find how many terms of the arithmetic progression 24, 21, 18, ..... must be

taken so that their sum is 78 using formula.

MID/T-25-26/28

7 81-E

31. Sides AB and AC and median AD of ∆ ABC are respectively proportional to

sides PQ and PR and median PM of another ∆ PQR. Show that

∆ ABC ~ ∆ PQR.

32. ABCD is a trapezium in which AB || DC and its diagonals intersect each

other at the point ‘O’. Show that
AO CO
= .
BO DO

33. The sum of the digits of a two digit number is 9. The number obtained by

reversing its digits is 9 less than twice the original number. Find the original

number.

V. Answer the following questions : 4 × 4 = 16

34. Find a quadratic polynomial whose sum and product of the zeroes are – 6

and 8 respectively. Find the zeroes of this polynomial. Verify the relationship

between the zeroes and the coefficients.

35. Solve the given pair of linear equations by graphical method :

x+y = 5

x + 2y = 6

36. The numerator of a fraction is 1 less than its denominator. If 3 is added to
3
both numerator and denominator, then the new fraction obtained is
28

more than the original fraction. Find the original fraction.

OR

MID/T-25-26/28 [ Turn over

81-E 8

There are three consecutive positive integers in ascending order such that

the sum of “the square of the first integer and the product of other two

integers is 29”. Find those integers.

37. Prove that “if in two triangles, corresponding angles are equal, then their

corresponding sides are in the same ratio ( or proportion ) and hence the two

triangles are similar”.

VI. Answer the following question : 1×5=5

38. In an arithmetic progression, the ratio of the 5 th term to the 24 th term is

1 : 5. If he sum of first 10 terms of this progression is 210 and the last term

of the progression is 99, then find the sum of all the terms of the arithmetic

progression.

MID/T-25-26/28

PÜ®ÝìoPÜ ÍÝÇÝ ±ÜÄàûæ ÊÜáñÜᤠÊÜåèƯ|ì¿á ÊÜáívÜÈ
6 ®æà AvÜxÃÜÓæ¤, ÊÜáÇæÉàÍÜÌÃÜí, ¸æíWÜÙÜãÃÜá & 560 003

KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
6th Cross, Malleshwaram, Bengaluru – 560 003
2025-26 ®æà ÓÝ騆 GÓ….GÓ….GÇ….Ô. A«ÜìÊÝÑìPÜ ±ÜÄàûæ
S.S.L.C. MID-TERM EXAMINATION FOR THE YEAR – 2025-26

ËÐÜ¿á : WÜ~ñÜ
Subject : MATHEMATICS
PܮܰvÜ ÊÜÞ«ÜÂÊÜá / Kannada Medium

ËÐÜ¿á ÓÜíPæàñÜ : 81-K Subject Code : 81-K
©®ÝíPÜ : 16. 09. 2025 ] [ Date : 16. 09. 2025
ÓÜÊÜá¿á ¸æÙÜWæY 10-30 Äí¨Ü ÊÜá«ÝÂÖÜ° 1-45 ÃÜÊÜÃæWæ ] [ Time : 10-30 A.M. to 1-45 P.M.
WÜÄÐÜu AíPÜWÜÙÜá 80 ] [ Max. Marks : 80

±ÜÄàûݦìWÝX ÓÝÊÜޮܠÓÜãaÜ®æWÜÙÜá
1. D ±ÜÅÍ氱ܣÅPæ¿áá Joár 38 ±ÜÅÍæ°WÜÙÜ®Üá° Öæãí©¨æ.

2. ±ÜÅÍæ°WÜÚWæ PæãqrÃÜáÊÜ ÓÜãaÜ®æWÜÙÜ®Üá° ±ÝÈÔ.

3. ŸÆ »ÝWܨÜÈÉ PæãqrÃÜáÊÜ AíQWÜÙÜá ¯ÜÅÍæ°WÜÚXÃÜáÊÜ ±Üä|ì AíPÜWÜÙÜ®Üá° ÓÜãbÓÜáñܤÊæ.

4. ±ÜÅÍ氱ܣÅPæ¿á®Üá° K©PæãÙÜÛÆá 15 ¯ËáÐÜWÜÙÜ PÝÇÝÊÜPÝÍÜÊÜâ ÓæàĨÜíñæ, EñܤÄÓÜÆá ¯WÜ©±ÜwÓÜÇݨÜ
ÓÜÊÜá¿áÊÜ®Üá° ±ÜÅÍ氱ܣÅPæ¿á ÊæáàÇݽWܨÜÈÉ ¯àvÜÇÝX¨æ.

I. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ A¥ÜÊÝ A±Üä|ì ÖæàÚPæWÜÚWæ ®ÝÆáR ±Ü¿Þì¿á EñܤÃÜWÜÙÜ®Üá° ¯àvÜÇÝX¨æ. AÊÜâWÜÙÜÈÉ
ÓÜãPܤÊÝ¨Ü EñܤÃÜÊÜ®Üá° BÄÔ, A¨ÜÃÜ PÜÅÊÜÞûÜÃܨæãvÜ®æ ±Üä|ì EñܤÃÜÊÜ®Üá° ŸÃæÀáÄ 8×1=8

1. p ( x ) = 2 x 3 + 5x – 6 D ŸÖÜá¯Ü¨æãàQ¤¿á ÊÜáÖÜñܤÊÜá [ÝñÜÊÜâ (wXÅ)

(A) 2 (B) 3

(C) 5 (D) 6

MID/T-25-26/27 [ Turn over

81-K 2
2. GÃÜvÜá aÜÃÝûÜÃÜÊÜâÙÜÛ
ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw a1x  b1y  c1  0 ÊÜáñÜá¤
a 2 x  b2y  c 2  0 CÊÜâWÜÙÜá A®Ü®Ü ±ÜÄÖÝÃÜÊÜ®Üá° Öæãí©¨ÜªÃæ, D PæÙÜX®ÜÊÜâWÜÙÜÈÉ ÓÜÄ¿Þ¨Ü
ْdž憉
a1 b1
(A) 
a2 b2

a1 b1 c1
(B)  
a2 b2 c2

a1 b1
(C) 
a2 b2

a1 b1 c1
(D)  
a2 b2 c2

3. D PæÙÜX®ÜÊÜâWÜÙÜÈÉ ÊÜWÜìÓÜËáàPÜÃÜ|ÊÜâ

(A) x 4  3x  2  0 (B) 2x + 3 = 0

(C) x 2  5x  6  0 (D) 2x 3  7x  1  0

4. 100, 93, 86 ..... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á ÓÝÊÜޮܠÊÜÂñÝÂÓÜÊÜâ

(A) – 7 (B) 7

(C) 3 (D) 5

5. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á n®æà ±Ü¨Ü 5n + 3 B¨ÝWÜ, ÍæÅà{¿á 3®æà ±Ü¨ÜÊÜâ
(A) 11 (B) 12

(C) 13 (D) 18

6. bñÜŨÜÈÉ XY || BC B¨ÜÃæ, AX Wæ ÓÜÊÜá®Ý¨Üá¨Üá
AB

AX AX
(A) (B)
AY XB
AY AC
(C) (D)
AC AY

MID/T-25-26/27

3 81-K
7. PæãqrÃÜáÊÜ ®Üûæ¿áÈÉ p ( x ) ŸÖÜá¯Ü¨æãàQ¤¿á ÍÜã¬ÜÂñæWÜÙÜ ÓÜíTæÂ

(A) 0 (B) 1

(C) 2 (D) 3

8. 2 x 2 – ( k + 1 ) x + 3 = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü Jí¨Üá ÊÜáãÆÊÜâ 1 B¨ÜÃæ, k ®Ü ¸æÇæ¿áá

(A) 0 (B) 1

(C) 3 (D) 4

II. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8×1=8

9. AíPÜWÜ~ñÜ¨Ü ÊÜáãÆ ±ÜÅÊæáà¿á ÊÜ®Üá° ¯ÃÜã²Ô.

10. GÃÜvÜá aÜÃÝûÜÃÜÊÜâÙÜÛ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿áá AÔ§ÃÜÊÝX¨ÜªÃæ, AÊÜâ GÐÜár ±ÜÄÖÝÃÜWÜÙÜ®Üá°
Öæãí©ÃÜáñܤÊæ

11. Êæã¨ÜÆ n «Ü®Ü ±ÜäOÝìíPÜWÜÙÜ ÊæãñܤÊܬÜá® PÜívÜá×w¿ááÊÜ ÓÜãñÜÅ ŸÃæÀáÄ.

12. ¸ÝÖÜáWÜÙÜ ÓÜíTæ Jí¨æà BXÃÜáÊÜ GÃÜvÜá ŸÖÜá»ÜáhÝPÜê£WÜÙÜ ÓÜÊÜáÃÜã¯ÜñæWæ CÃÜáÊÜ GÃÜvÜá
­Ÿí«Ü¬æWÜÙܬÜá® ŸÃæÀáÄ.

13. p ( x ) = 2 x 2 – 3x + k D ŸÖÜá¯Ü¨æãàQ¤¿á ÍÜã¬ÜÂñæWÜÙÜ WÜá|ÆŸœÊÜâ 3 B¨ÜÃæ, k ¿á
¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

MID/T-25-26/27 [ Turn over

81-K 4
14. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á ÊÜáãÃÜá PÜÅÊÜÞ®ÜáWÜñÜ ±Ü¨ÜWÜÙÜ Êæãñܤ 21 B¨ÜÃæ, GÃÜvÜ®æà ±Ü¨ÜÊÜ®Üá°
PÜívÜá×wÀáÄ.

15. 3 ¸ÝÂp… ÊÜáñÜᤠ2 aæívÜáWÜÙÜ Joár ¸æÇæ¿áá ÃÜã. 700 BX¨æ. Jí¨Üá aæíw®Ü ¸æÇæ ÃÜã. 50 B¨ÜÃæ,
Jí¨Üá ¸ÝÂp…¬Ü ¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

AB 2
16.  ABC ~  DEF BX¨æ.  ÊÜáñÜᤠBC = 4 cm B¨ÜÃæ, EF AÙÜñæ¿á®Üá°
DE 3
PÜívÜá×wÀáÄ.

III. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 8 × 2 = 16

17. ÓÜí¿ááPܤ ÓÜíTæ Gí¨ÜÃæà®Üá 23 ÊÜáñÜᤠ24 CÊÜâWÜÙÜÈÉ ÓÜí¿ááPܤ ÓÜíTæ ¿ÞÊÜâ¨Üá

18. 2  5 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ Gí¨Üá ÓݘÔ.

19. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿á®Üá° ÊÜiìÓÜáÊÜ Ë«Ý®Ü©í¨Ü ¹wÔ

2x + y = 8

3x – y = 7

A¥ÜÊÝ

a1 b1 c1
, ÊÜáñÜᤠA®Üá±ÝñÜWÜÙÜ®Üá° ÖæãàÈÓÜáÊÜ ÊÜáãÆPÜ,
a2 b2 c2

9x + 3y + 12 = 0 ÊÜáñÜᤠ18x + 6y + 24 = 0 ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàw¿áá, Jí¨Üá
¹í¨ÜáˬÜÈÉ dæà©ÓÜáñܤÊæÁáà A¥ÜÊÝ ÓÜÊÜÞíñÜÃÜÊÝXÊæÁáà A¥ÜÊÝ IPÜÂWæãíwÊæÁáà GíŸá¨Ü¬Üá®
PÜívÜá×wÀáÄ.

20. 1, 5, 9, 13, ....... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á ±Ü¨ÜÊÜâ 77 BX¨æÁáà GíŸá¨Ü¬Üá® ÓÜãñÜÅ ŸÙÜÔ
PÜívÜá×wÀáÄ.

A¥ÜÊÝ

Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á ÓÝÊÜޮܠÊÜÂñÝÂÓÜ 3 ÊÜáñÜᤠA¨ÜÃÜ ÖÜñܤ®æà ±Ü¨Ü 32 BX¨æ. ÖÝWݨÜÃæ B
ÍæÅà{¿á Êæã¨ÜÆ®æ¿á ±Ü¨ÜÊÜ®Üá° PÜívÜá×wÀáÄ.
MID/T-25-26/27

5 81-K
21. PæãqrÃÜáÊÜ bñÜŨÜÈÉ DE || OQ ÊÜáñÜᤠDF || OR BX¨æ. ÖÝWݨÜÃæ EF || QR Gí¨Üá ñæãàÄÔ.

22.  ABC ¿áÈÉ DE || BC BX¨æ. AD = x, DB = x – 4, AE = x + 4 ÊÜáñÜᤠEC = x – 2
B¨ÜÃæ, x ®Ü ¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

23. JŸº ÎûÜPÜÃÜ ŸÚ 96 ¯æ¬… ÊÜáñÜᤠ128 ¯æ­ÕÇ…WÜÚÊæ. ¿ÞÊÜâ¨æà ¯æ¬… ÊÜáñÜᤠ¯æ­ÕÇ…WÜÙÜá EÚ¿á¨Üíñæ,
WÜÄÐÜu GÐÜár ˨ݦìWÜÚWæ ÓÜÊÜá¬ÝX ¯æ¬… ÊÜáñÜᤠ¯æ­ÕÇ…WÜÙܬÜá® ËñÜÃÜOæ ÊÜÞvÜŸÖÜá¨Üá Gí¨Üá
PÜívÜá×wÀáÄ.

24. A ÊÜáñÜᤠB GíŸ CŸºÃÜá ˨ݦìWÜÙÜ ŸÚ Joár 25 WæãàÈWÜÚÊæ. ¯ÜÅ£Á㟺ÃÜã 5 WæãàÈWÜÙÜ®Üá°
PÜÙæ¨ÜáPæãívÜÃÜá. DWÜ AÊÜÃÜ ŸÚ EÚ¿ááÊÜ WæãàÈWÜÙÜ WÜá|ÆŸœÊÜâ 50 BWÜáñܤ¨æ. AÊÜÃÜ ŸÚ C¨Üª
WæãàÈWÜÙÜ ÓÜíTæ¿á®Üá° PÜívÜá×w¿áÆá ÊÜWÜìÓÜËáàPÜÃÜ|ÊÜ®Üá° B¨ÜÍÜìÃÜã±Ü¨ÜÈÉ ÃÜbÔ.

IV. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 9 × 3 = 27

25. 3 Jí¨Üá A»ÝWÜÆŸœ ÓÜíTæ Gí¨Üá ÓݘÔ.

A¥ÜÊÝ

510 ÊÜáñÜᤠ92 D ÓÜíTæÂWÜÙÜ Æ.ÓÝ.A. ÊÜáñÜᤠÊÜá.ÓÝ.A.WÜÙÜ®Üá° AË»Ýg A±ÜÊÜñÜì®Ü ˫ݮܩí¨Ü
PÜívÜá×wÀáÄ ÊÜáñÜᤠB ÓÜíTæÂWÜÙÜ Æ.ÓÝ.A.  ÊÜá.ÓÝ.A. = ÓÜíTæÂWÜÙÜ WÜá|ÆŸœ GíŸá¨Ü¬Üá® ñÝÙæ
®æãàw.

MID/T-25-26/27 [ Turn over

81-K 6
26. p ( x ) = x 2 + 3x + 1 D ŸÖÜá¯Ü¨æãàQ¤¿á GÃÜvÜá ÍÜã¬ÜÂñæWÜÙÜá  ÊÜáñÜᤠ B¨ÜÃæ, D PæÙÜX®ÜÊÜâWÜÙÜ
¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

1 1
i) 
 

ii)  2  2 .

A¥ÜÊÝ

p ( x ) = 2 x 2 – 6x + k D ŸÖÜá¯Ü¨æãàQ¤¿á Jí¨Üá ÍÜã¬ÜÂñæ¿áá C¬æã®í¨Üá ÍÜã¬ÜÂñæ¿á
GÃÜvÜÃÜÐÜár B¨ÝWÜ, k ¿á ¸æÇæ¿á®Üá° PÜívÜá×wÀáÄ.

27. Jí¨Üá ÃæçÆá 480 km ¨ÜãÃÜÊÜ®Üá° HPÜÃÜã±Ü gÊܨæãí©Wæ PÜÅËáÓÜáñܤ¨æ. A¨ÜÃÜ gÊÜÊÜâ 10 km/h
A˜PÜÊݨÝWÜ, AÐærà ¨ÜãÃÜÊܬÜá® PÜÅËáÓÜÆá 4 WÜípæ PÜwÊæá ñæWæ¨ÜáPæãÙÜáÛñܤ¨æ. ÖÝWݨÜÃæ Ãæç騆 gÊÜÊÜ®Üá°
PÜívÜá×wÀáÄ.

A¥ÜÊÝ

Jí¨Üá ÆíŸPæãà¬Ü £Å»Üág¨Ü GñܤÃÜÊÜâ A¨ÜÃÜ ¯Ý¨ÜQRíñÜ 7 cm PÜwÊæá C¨æ. A¨ÜÃÜ ËPÜ|ì¨Ü E¨ÜªÊÜâ
13 cm B¨ÜÃæ, EÚ¨æÃÜvÜá ¸ÝÖÜáWÜÙÜ E¨ÜªWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

28. x 2 + 7x + 12 = 0 D ÊÜWÜìÓÜËáàPÜÃÜ|¨Ü Íæãà«ÜPÜÊÜ®Üá° PÜívÜá×wÀáÄ. D ÓÜËáàPÜÃÜ|ÊÜâ ÊÝÓܤÊÜ
ÊÜáãÆWÜÙÜ®Üá° Öæãí©¨ÜªÃæ, AÊÜâWÜÙÜ®Üá° PÜívÜá×wÀáÄ.

29. 5 ÊÜÐÜìWÜÙÜ ŸÚPÜ A ®Ü ÊÜ¿áÓÜáÕ AÊÜÃÜ ÊÜáWÜ®Ü ÊÜ¿áÔÕ®Ü ÊÜáãÃÜÃÜÐÜár BWÜáñܤ¨æ. 5 ÊÜÐÜìWÜÙÜ ×í¨æ
A ®Ü ÊÜ¿áÓÜáÕ AÊÜÃÜ ÊÜáWÜ¬Ü ÊÜ¿áÔÕ¬Ü HÙÜÃÜÑr¨ÜªÃæ, AÊÜÄŸºÃÜ DX¬Ü ÊÜ¿áÓÜáÕWÜÙܬÜá® PÜívÜá×wÀáÄ.

30. 5 Äí¨Ü »ÝXÔ¨ÝWÜ 1 ®Üá° ÍæàÐÜÊÝX Öæãí©ÃÜáÊÜ, 1 ÊÜáñÜᤠ100 ÃÜ ®ÜvÜá訆 GÇÝÉ ±ÜäOÝìíPÜWÜÙÜ
ÊæãñܤÊÜ®Üá° ÓÜãñÜÅ E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

A¥ÜÊÝ

24, 21, 18, ..... D ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á GÐÜár ±Ü¨ÜWÜÙÜ Êæãñܤ 78 BWÜáñܤ¨æ, GíŸá¨Ü¬Üá® ÓÜãñÜÅ
E±ÜÁãàXÔ PÜívÜá×wÀáÄ.

MID/T-25-26/27

7 81-K
31.  ABC ¿á ¸ÝÖÜáWÜÙÝ¨Ü AB ÊÜáñÜᤠAC WÜÙÜá ÖÝWÜã ÊÜá«ÜÂÃæàTæ AD ¿áá PÜÅÊÜáÊÝX  PQR ®Ü
¸ÝÖÜáWÜÙÝ¨Ü PQ ÊÜáñÜᤠPR ÖÝWÜã ÊÜá«ÜÂÃæàTæ PM ®æãí©Wæ ÓÜÊÜÞ®Üá±ÝñÜ Öæãí©¨ÜªÃæ,
 ABC ~  PQR Gí¨Üá ñæãàÄÔ.

32. ABCD Jí¨Üá ñÝŲgÂ. C¨ÜÃÜÈÉ AB || DC ÊÜáñÜᤠPÜ|ìWÜÙÜá ±ÜÃÜÓܳÃÜ ‘O’ ¹í¨ÜáË®ÜÈÉ dæà©ÓÜáñܤÊæ.

ÖÝWݨÜÃæ, AO 
CO
Gí¨Üá ñæãàÄÔ.
BO DO

33. GÃÜvÜíQ¿á Jí¨Üá ÓÜíTæ¿á AíQWÜÙÜ Êæãñܤ 9 BX¨æ. D ÓÜíTæ¿á AíQWÜÙܬÜá® A¨ÜÆá Ÿ¨ÜÆá
ÊÜÞw¨ÝWÜ ¨æãÃæ¿ááÊÜ ÓÜíTæ¿áá ÊÜáãÆ ÓÜíTæ¿á GÃÜvÜÃÜÐÜrQRíñÜ 9 PÜwÊæá C¨æ. ÊÜáãÆ
ÓÜíTæ¿á®Üá° PÜívÜá×wÀáÄ.

V. PæÙÜX®Ü ±ÜÅÍæ°WÜÚWæ EñܤÄÔ 4 × 4 = 16

34. ÍÜã¬ÜÂñæWÜÙÜ Êæãñܤ ÊÜáñÜᤠWÜá|ÆŸœWÜÙÜá PÜÅÊÜáÊÝX – 6 ÊÜáñÜᤠ8 BXÃÜáÊÜ ÊÜWÜìŸÖÜá¯Ü¨æãàQ¤¿á¬Üá®
PÜívÜá×wÀáÄ. D ÊÜWÜìŸÖÜá¯Ü¨æãàQ¤¿á ÍÜã¬ÜÂñæWÜÙܬÜá® PÜívÜá×w¨Üá, ÍÜã¬ÜÂñæWÜÙÜá ÊÜáñÜá¤
ÓÜÖÜWÜá|PÜWÜÙÜ ¬ÜvÜáË¬Ü ÓÜíŸí«ÜÊܬÜá® ñÝÙæ ®æãàw.

35. PæãqrÃÜáÊÜ ÃæàTÝñܾPÜ ÓÜËáàPÜÃÜ|WÜÙÜ hæãàwWæ ®Üûæ¿á ˫ݮܩí¨Ü ±ÜÄÖÝÃÜÊÜ®Üá° PÜívÜá×wÀáÄ

x+y = 5

x + 2y = 6

MID/T-25-26/27 [ Turn over

81-K 8
36. Jí¨Üá ¼®Ü°ÃÝοá AíÍÜÊÜâ A¨ÜÃÜ dæà¨ÜQRíñÜ 1 PÜwÊæá C¨æ. AíÍÜ ÊÜáñÜᤠdæà¨Ü CÊæÃÜvÜPÜãR 3 ®Üá°
3
PÜãw¨ÝWÜ EípÝWÜáÊÜ ÖæãÓÜ ¼®Ü°ÃÝοáá ÊÜáãÆ ¼®Ü°ÃÝÎXíñÜ ÖæaÝcX¨æ. ÊÜáãÆ
28
¼®Ü°ÃÝοá®Üá° PÜívÜá×wÀáÄ.

A¥ÜÊÝ

HÄPæ PÜÅÊÜá¨ÜÈÉÃÜáÊÜ ÊÜáãÃÜá PÜÅÊÜÞ®ÜáWÜñÜ «Ü®Ü ±ÜäOÝìíPÜWÜÙÜÈÉ Êæã¨ÜÆ ±ÜäOÝìíPÜ¨Ü ÊÜWÜì ÊÜáñÜá¤
EÚ¨æÃÜvÜá ¯ÜäOÝìíPÜWÜÙÜ WÜá|ÆŸœ CÊÜâWÜÙÜ ÊæãñܤÊÜâ 29 BX¨æ. B ±ÜäOÝìíPÜWÜÙÜ®Üá°
PÜívÜá×wÀáÄ.

37. GÃÜvÜá £Å»ÜágWÜÙÜÈÉ A®ÜáÃÜã±Ü Pæãà®ÜWÜÙÜá ÓÜÊÜáÊݨÜÃæ AÊÜâWÜÙÜ A®ÜáÃÜã±Ü ¸ÝÖÜáWÜÙÜ A®Üá±ÝñÜWÜÙÜá
ÓÜÊÜá A¥ÜÊÝ ÓÜÊÜÞ®Üá±ÝñܨÜÈÉÃÜáñܤ¨æ B¨ÜªÄí¨Ü B £Å»ÜágWÜÙÜá ÓÜÊÜáÃÜã±ÜÊÝXÃÜáñܤÊæ Gí¨Üá
ÓݘÔ.

VI. PæÙÜX®Ü ±ÜÅÍæ°Wæ EñܤÄÔ 1×5=5

38. Jí¨Üá ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿áÈÉ, 5 ®æà ±Ü¨Ü ÊÜáñÜᤠ24 ®æà ±Ü¨ÜWÜÙÜ A®Üá±ÝñÜ 1 : 5 BX¨æ. D ÍæÅà{¿á
Êæã¨ÜÆ 10 ±Ü¨ÜWÜÙÜ Êæãñܤ 210 ÊÜáñÜᤠÍæÅà{¿á Pæã®æà ±Ü¨ÜÊÜâ 99 B¨ÜÃæ, ÓÜÊÜÞíñÜÃÜ ÍæÅà{¿á GÇÝÉ
±Ü¨ÜWÜÙÜ ÊæãñܤÊÜ®Üá° PÜívÜá×wÀáÄ.

MID/T-25-26/27

AglaSem Earn While You Learn Program!

Share your recent question papers with us and get paid!
अपने हाल के न प हम भेज और कमाएँ!

To participate, email us with the following details:
भाग लेने के लए हम न न ववरण के साथ ईमेल कर:

- Your Name
- Your Class
- Your Board
- Details of the Question Papers you have
आपके पास उपल ध न प का ववरण

Our team will reach out to you with the next steps.
हमार ट म आगे क या के लए आपसे संपक करे गी।

Email: support@

Study Materials
Notes

Model Papers Class 6 Notes

Sample Papers Class 7 Notes
Half Yearly Sample Papers Class 8 Notes

Class 9 Notes
Important Resources
Class 10 Notes
Periodic Table
Class 11 Notes
Writing Skills / Formats

Maps of India / World Class 12 Notes

Books and Solutions

NCERT Books
NCERT Book Solutions
HC Verma Chapter Wise Solutions
RD Sharma Solutions
CGBSE Solutions